text:
## Physics

### Implementing Fault-tolerant Entangling Gates on the Five-qubit Code and the Color Code. (arXiv:2208.01863v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We compare two different implementations of fault-tolerant entangling gates on logical qubits. In one instance, a twelve-qubit trapped-ion quantum computer is used to implement a non-transversal logical CNOT gate between two five qubit codes. The operation is evaluated with varying degrees of fault tolerance, which are provided by including quantum error correction circuit primitives known as flagging and pieceable fault tolerance. In the second instance, a twenty-qubit trapped-ion quantum computer is used to implement a transversal logical CNOT gate on two [[7,1,3]] color codes. The two codes were implemented on different but similar devices, and in both instances, all of the quantum error correction primitives, including the determination of corrections via decoding, are implemented during runtime using a classical compute environment that is tightly integrated with the quantum processor. For different combinations of the primitives, logical state fidelity measurements are made after applying the gate to different input states, providing bounds on the process fidelity. We find the highest fidelity operations with the color code, with the fault-tolerant SPAM operation achieving fidelities of 0.99939(15) and 0.99959(13) when preparing eigenstates of the logical X and Z operators, which is higher than the average physical qubit SPAM fidelities of 0.9968(2) and 0.9970(1) for the physical X and Z bases, respectively. When combined with a logical transversal CNOT gate, we find the color code to perform the sequence--state preparation, CNOT, measure out--with an average fidelity bounded by [0.9957,0.9963]. The logical fidelity bounds are higher than the analogous physical-level fidelity bounds, which we find to be [0.9850,0.9903], reflecting multiple physical noise sources such as SPAM errors for two qubits, several single-qubit gates, a two-qubit gate and some amount of memory error.

Categories: Journals, Physics

### Enhancing spin squeezing using soft-core interactions. (arXiv:2208.01869v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We propose a new protocol for preparing spin squeezed states in controllable atomic, molecular, and optical systems, with particular relevance to emerging optical clock platforms compatible with Rydberg interactions. By combining a short-ranged, soft-core potential with an external drive, we can transform naturally emerging Ising interactions into an XX spin model while opening a many-body gap. The gap helps maintain the system within a collective manifold of states where metrologically useful spin squeezing can be generated at a level comparable to the spin squeezing generated in systems with genuine all-to-all interactions. We examine the robustness of our protocol to experimentally-relevant decoherence and show favorable performance over typical protocols lacking gap protection.

Categories: Journals, Physics

### Photon-mediated correlated hopping in a synthetic ladder. (arXiv:2208.01896v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We propose a new direction in quantum simulation that uses multilevel atoms in an optical cavity as a toolbox to engineer new types of bosonic models featuring correlated hopping processes in a synthetic ladder spanned by atomic ground states. The underlying mechanisms responsible for correlated hopping are collective cavity-mediated interactions that dress a manifold of excited levels in the far detuned limit. By weakly coupling the ground state levels to these dressed states using two laser drives with appropriate detunings, one can engineer correlated hopping processes while suppressing undesired single-particle and collective shifts of the ground state levels. We discuss the rich many-body dynamics that can be realized in the synthetic ladder including pair production processes, chiral transport and light-cone correlation spreading. The latter illustrates that an effective notion of locality can be engineered in a system with fully collective interactions.

Categories: Journals, Physics

### Semiquantum secret sharing by using x-type states. (arXiv:2208.01922v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

In this paper, a semiquantum secret sharing (SQSS) protocol based on x-type states is proposed, which can accomplish the goal that only when two classical communicants cooperate together can they extract the shared secret key of a quantum communicant. Detailed security analysis turns out that this protocol can resist the participant attack and the outside attack. This protocol has some merits: (1) it only requires one kind of quantum entangled state as the initial quantum resource; (2) it doesn't employ quantum entanglement swapping or unitary operations; and (3) it needn't share private keys among different participants beforehand.

Categories: Journals, Physics

### Comparing the quantum switch and its simulations with energetically-constrained operations. (arXiv:2208.01952v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Quantum mechanics allows processes to be superposed, leading to a genuinely quantum lack of causal structure. For example, the process known as the quantum switch applies two operations ${\cal A}$ and ${\cal B}$ in a superposition of the two possible orders, ${\cal A}$ before ${\cal B}$ and ${\cal B}$ before ${\cal A}$. Experimental implementations of the quantum switch have been challenged by some on the grounds that the operations ${\cal A}$ and ${\cal B}$ were implemented more than once, thereby simulating indefinite causal order rather than actually implementing it. Motivated by this debate, we consider a situation in which the quantum operations are physically described by a light-matter interaction model. When one restricts the energy available for the implementations, an imperfect operation creating correlations between a "target" system and its environment is implemented instead, allowing one to distinguish processes using different numbers of operations. We consider such an energetically-constrained scenario and compare the quantum switch to one of its natural simulations, where each operation is implemented twice. Considering a commuting-vs-anticommuting unitary discrimination task, we find that within our model the quantum switch performs better, for some fixed amount of energy, than its simulation. In addition to the known computational or communication advantages of causal superpositions, our work raises new questions about their potential energetic advantages.

Categories: Journals, Physics

### Fingerprint and universal Markovian closure of structured bosonic environments. (arXiv:2208.01978v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We exploit the properties of chain mapping transformations of bosonic environments to identify a finite collection of modes able to capture the characteristic features, or fingerprint, of the environment. Moreover we show that the countable infinity of residual bath modes can be replaced by a universal Markovian closure, namely a small collection of damped modes undergoing a Lindblad-type dynamics whose parametrization is independent of the spectral density under consideration. We show that the Markovian closure provides a quadratic speed-up with respect to standard chain mapping techniques and makes the memory requirement independent of the simulation time, while preserving all the information on the fingerprint modes. We illustrate the application of the Markovian closure to the computation of linear spectra but also to non-linear spectral response, a relevant experimentally accessible many body coherence witness for which efficient numerically exact calculations in realistic environments are currently lacking.

Categories: Journals, Physics

### All-optical coherent quantum-noise cancellation in cascaded optomechanical systems. (arXiv:2208.01982v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Coherent quantum noise cancellation (CQNC) can be used in optomechanical sensors to surpass the standard quantum limit (SQL). In this paper, we investigate an optomechanical force sensor that uses the CQNC strategy by cascading the optomechanical system with an all-optical effective negative mass oscillator. Specifically, we analyze matching conditions, losses and compare the two possible arrangements in which either the optomechanical or the negative mass system couples first to light. While both of these orderings yield a sub-SQL performance, we find that placing the effective negative mass oscillator before the optomechanical sensor will always be advantageous for realistic parameters. The modular design of the cascaded scheme allows for better control of the sub-systems by avoiding undesirable coupling between system components, while maintaining similar performance to the integrated configuration proposed earlier. We conclude our work with a case study of a micro-optomechanical implementation.

Categories: Journals, Physics

### High validity entanglement verification with finite copies of a quantum state. (arXiv:2208.01983v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Detecting entanglement of multipartite quantum states is an inherently probabilistic process due to a finite number of measured samples. The level of confidence of entanglement detection can be used to quantify the probability that the measured signal is coming from a separable state and provides a meaningful figure of merit for big data sets. Yet, for limited sample sizes, to avoid serious misinterpretations of the experimental results, one should not only consider the probability that a separable state gave rise to the measured signal, but should also include information about the probability that the signal came from an entangled state. We demonstrate this explicitly and propose a comprehensive method of entanglement detection when only a very limited amount of data is available. The method is based on a non-linear combination of correlation functions and is independent of system size. As an example, we derive the optimal number of measurement settings and clicks per setting revealing entanglement with only 20 copies of a state.

Categories: Journals, Physics

### Disappearance and reappearance of above-threshold-ionization peaks. (arXiv:2208.02039v1 [physics.atom-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

It is shown that above-threshold ionization peaks disappear when the kinetic energy associated with the nondipole radiation-pressure-induced photoelectron momentum in the laser propagation direction becomes comparable to the photon energy, and how peaks can be made reappear if knowledge of the length and direction of the photoelectron momentum is at hand and an emission-direction-dependent momentum shift is accounted for. The reported findings should be observable with intense mid-infrared laser pulses.

Categories: Journals, Physics

### Quantum-Assisted Greedy Algorithms. (arXiv:2208.02042v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We show how to leverage quantum annealers (QAs) to better select candidates in greedy algorithms. Unlike conventional greedy algorithms that employ problem-specific heuristics for making locally optimal choices at each stage, we use QAs that sample from the ground state of problem-dependent Hamiltonians at cryogenic temperatures and use retrieved samples to estimate the probability distribution of problem variables. More specifically, we look at each spin of the Ising model as a random variable and contract all problem variables whose corresponding uncertainties are negligible. Our empirical results on a D-Wave 2000Q quantum processor demonstrate that the proposed quantum-assisted greedy algorithm (QAGA) scheme can find notably better solutions compared to the state-of-the-art techniques in the realm of quantum annealing

Categories: Journals, Physics

### Quantum Approximate Optimization Algorithm in Non-Markovian Quantum Systems. (arXiv:2208.02066v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Quantum Approximate Optimization Algorithm(QAOA) is a promising quantum algorithm that can demonstrate quantum supremacy. The performance of QAOA on noisy intermediate-scale quantum (NISQ) devices degrades due to decoherence. In this paper, we present a framework for running QAOA on non-Markovian quantum systems which are represented by an augmented system model. In this model, a non-Markovian environment is modelled as an ancillary system driven by quantum white noises and the corresponding principal system is the computational unit for the algorithm. With this model, we mathematically formulates QAOA as piecewise control of the augmented system. To reduce the effect of non-Markovian decoherence, the above basic algorithm is modified for obtaining an efficient depth by a proximal gradient descent algorithm. Finally, in an example of the Max-Cut problem, we find non-Markovianity can help to achieve a good performance of QAOA, which is characterized by an exploration rate.

Categories: Journals, Physics

### Entanglement spectrum and entropy in Floquet topological matter. (arXiv:2208.02075v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Entanglement is one is the most fundamental nature of quantum systems. In this work, we formulate the entanglement spectrum and entropy for Floquet noninteracting fermionic lattice models and build their connections with Floquet topological phases. Topological winding and Chern numbers are introduced to characterize the entanglement spectrum and eigenmodes in one and two spatial dimensions. Correspondences between the spectrum and topology of Floquet entanglement Hamiltonians under periodic boundary conditions and topological edge states under open boundary conditions are further established for Floquet topological insulators in different symmetry classes and spatial dimensions. Our study provides a firm entry to explore richer entanglement patterns in Floquet topological matter.

Categories: Journals, Physics

### Optimizing Counterdiabaticity by Variational Quantum Circuits. (arXiv:2208.02087v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Utilizing counterdiabatic (CD) driving - aiming at suppression of diabatic transition - in digitized adiabatic evolution have garnered immense interest in quantum protocols and algorithms. However, improving the approximate CD terms with a nested commutator ansatz is a challenging task. In this work, we propose a technique of finding optimal coefficients of the CD terms using a variational quantum circuit. By classical optimizations routines, the parameters of this circuit are optimized to provide the coefficients corresponding to the CD terms. Then their improved performance is exemplified in Greenberger-Horne-Zeilinger state preparation on nearest-neighbor Ising model. Finally, we also show the advantage over the usual quantum approximation optimization algorithm, in terms of fidelity with bounded time.

Categories: Journals, Physics

### Mapping atomic trapping in an optical superlattice onto the libration of a planar rotor in electric fields. (arXiv:2208.02096v1 [cond-mat.quant-gas])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We show that two seemingly unrelated problems - the trapping of an atom in an optical superlattice (OSL) and the libration of a planar rigid rotor in combined electric and optical fields - have isomorphic Hamiltonians. Formed by the interference of optical lattices whose spatial periods differ by a factor of two, OSL gives rise to a periodic potential that acts on atomic translation via the AC Stark effect. The latter system, also known as the generalized planar pendulum (GPP), is realized by subjecting a planar rigid rotor to combined orienting and aligning interactions due to the coupling of the rotor's permanent and induced electric dipole moments with the combined fields. The mapping makes it possible to establish correspondence between concepts developed for the two eigenproblems individually, such as localization on the one hand and orientation/alignment on the other. Moreover, since the GPP problem is conditionally quasi-exactly solvable (C-QES), so is atomic trapping in an OSL. We make use of both the correspondence and the quasi-exact solvability to treat ultracold atoms in an optical superlattice as a semifinite-gap system. The band structure of this system follows from the eigenenergies and their genuine and avoided crossings obtained previously for the GPP as analytic solutions of the Whittaker-Hill equation. These solutions characterize both the squeezing and the tunneling of atoms trapped in an optical superlattice and pave the way to unraveling their dynamics in analytic form.

Categories: Journals, Physics

### An apparatus for in-vacuum loading of nanoparticles into an optical trap. (arXiv:2208.02102v1 [physics.optics])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

We describe the design, construction, and operation of an apparatus utilizing a piezoelectric transducer for in-vacuum loading of nanoparticles into an optical trap for use in levitated optomechanics experiments. In contrast to commonly used nebulizer-based trap-loading methods which generate aerosolized liquid droplets containing nanoparticles, the method produces dry aerosols of both spherical and high-aspect ratio particles ranging in size by approximately two orders of mangitude. The device has been shown to generate accelerations of order $10^7$ $g$, which is sufficient to overcome stiction forces between glass nanoparticles and a glass substrate for particles as small as $170$ nm diameter. Particles with sizes ranging from $170$ nm to $\sim 10$ $\mu$m have been successfully loaded into optical traps at pressures ranging from $1$ bar to $0.6$ mbar. We report the velocity distribution of the particles launched from the substrate and our results indicate promise for direct loading into ultra-high-vacuum with sufficient laser feedback cooling. This loading technique could be useful for the development of compact fieldable sensors based on optically levitated nanoparticles as well as matter-wave interference experiments with ultra-cold nano-objects which rely on multiple repeated free-fall measurements and thus require rapid trap re-loading in high vacuum conditions.

Categories: Journals, Physics

### Single-cycle THz-field electro-optical sampling with single-photon detectors. (arXiv:2208.02103v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Terahertz time-domain spectroscopy (THz-TDS) using electro-optic sampling and ultrashort pulsed probes is a well-established technique for directly measuring the electric field of THz radiation. Traditionally, a balanced detection scheme relies on measuring an optical phase shift brought by THz-induced birefringence radiation using photodiodes, where sensitivity is limited by the shot-noise of the optical-sampling probe. Improvement to the sensitivity of such an approach could be achieved by applying quantum metrology, such as using NOON states for Heisenberg-limited phase estimation. We report on the first step in that direction, demonstrating that THz electric fields can be measured with single-photon detectors using a squeezed vacuum as the optical probe. Our approach achieves THz electro-optical sampling using phase-locked single-photon detectors at the shot-noise limit and thus paves the way toward quantum-enhanced THz sensing.

Categories: Journals, Physics

### Active Learning on a Programmable Photonic Quantum Processor. (arXiv:2208.02104v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Training a quantum machine learning model generally requires a large labeled dataset, which incurs high labeling and computational costs. To reduce such costs, a selective training strategy, called active learning (AL), chooses only a subset of the original dataset to learn while maintaining the trained model's performance. Here, we design and implement two AL-enpowered variational quantum classifiers, to investigate the potential applications and effectiveness of AL in quantum machine learning. Firstly, we build a programmable free-space photonic quantum processor, which enables the programmed implementation of various hybrid quantum-classical computing algorithms. Then, we code the designed variational quantum classifier with AL into the quantum processor, and execute comparative tests for the classifiers with and without the AL strategy. The results validate the great advantage of AL in quantum machine learning, as it saves at most $85\%$ labeling efforts and $91.6\%$ percent computational efforts compared to the training without AL on a data classification task. Our results inspire AL's further applications in large-scale quantum machine learning to drastically reduce training data and speed up training, underpinning the exploration of practical quantum advantages in quantum physics or real-world applications.

Categories: Journals, Physics

### A Simple and Tighter Derivation of Achievability for Classical Communication over Quantum Channels. (arXiv:2208.02132v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Achievability in information theory refers to demonstrating a coding strategy that accomplishes a prescribed performance benchmark for the underlying task. In quantum information theory, the crafted Hayashi-Nagaoka operator inequality is an essential technique in proving a wealth of one-shot achievability bounds since it effectively resembles a union bound in various problems. In this work, we show that the pretty-good measurement naturally plays a role as the union bound as well. A judicious application of it considerably simplifies the derivation of one-shot achievability for classical-quantum (c-q) channel coding via an elegant three-line proof.

The proposed analysis enjoys the following favorable features: (i) The established one-shot bound admits a closed-form expression as in the celebrated Holevo-Helstrom Theorem. Namely, the average error probability of sending $M$ messages through a c-q channel is upper bounded by the error of distinguishing the joint state between channel input and output against $(M-1)$-many products of its marginals. (ii) Our bound directly yields asymptotic results in the large deviation, small deviation, and moderate deviation regimes in a unified manner. (iii) The coefficients incurred in applying the Hayashi-Nagaoka operator inequality are no longer needed. Hence, the derived one-shot bound sharpens existing results that rely on the Hayashi-Nagaoka operator inequality. In particular, we obtain the tightest achievable $\epsilon$-one-shot capacity for c-q channel heretofore, and it improves the third-order coding rate in the asymptotic scenario. (iv) Our result holds for infinite-dimensional Hilbert space. (v) The proposed method applies to deriving one-shot bounds for data compression with quantum side information, entanglement-assisted classical communication over quantum channels, and various quantum network information-processing protocols.

Categories: Journals, Physics

### Quantum mean centering for block-encoding-based quantum algorithm. (arXiv:2208.02143v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

Mean centering is an important data preprocessing technique, which has a wide range of applications in data mining, machine learning, and multivariate statistical analysis. When the data set is large, this process will be time-consuming. In this paper, we propose an efficient quantum mean centering based on the block-encoding technique, which makes the existing quantum algorithms can get rid of the assumption that the original data set has been classically mean-centered. Specifically, we first introduce the centring matrix $C$ and propose in detail how to construct the block-encoding of $C$, and further obtain the block-encodings of $XC$, $CX$ and $CXC$ which represent removing the row means, column means and row-column means of the original data matrix $X$, respectively. Finally, combined with the quantum technology of block-encoding, our algorithm is successfully applied to principal component analysis, linear discriminant analysis and other algorithms. It is worth emphasizing that our algorithm also can be used in many algorithms involving matrix algebra problems, such as multidimensional scaling, kernel machine learning, and extreme learning machine.

Categories: Journals, Physics

### Entropic Dynamics and Quantum "Measurement". (arXiv:2208.02156v1 [quant-ph])

arXiv.org: Quantum Physics - Thu, 2022-08-04 22:45

The entropic dynamics (ED) approach to quantum mechanics is ideally suited to address the problem of measurement because it is based on entropic and Bayesian methods of inference that have been designed to process information and data. The approach succeeds because ED achieves a clear-cut separation between ontic and epistemic elements: positions are ontic while probabilities and wave functions are epistemic. Thus, ED is a viable realist psi-epistemic model. Such models are widely assumed to be ruled out by various no-go theorems. We show that ED evades those theorems by adopting a purely epistemic dynamics and denying the existence of an ontic dynamics at the subquantum level.

Categories: Journals, Physics


text:
# Is there a “\nocolumnbreak” analog to \nopagebreak?

If I want to prevent latex from putting two (short) paragraphs on different columns, is there something similar to \nopagebreak that will let me do this?

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AFAIK columns are handled like different pages and so \nopagebreak should do what you want. – Martin Scharrer May 20 '11 at 9:21
@Martin I couldn't get \nopagebreak to work, but maybe I just didn't insist on it enough. – Viktor Dahl May 20 '11 at 9:38
You could place the paragraphs into a {minipage}{\columnwidth} environment to prevent page or column breaking between or inside them. – Martin Scharrer May 20 '11 at 9:42
@Martin Thanks! But shouldn't you last comment be an answer? – Viktor Dahl May 20 '11 at 11:04

AFAIK columns are handled like different pages and so \nopagebreak should do what you want.
However, could simply wrap the paragraphs into \begin{minipage}{\columnwidth} ... \end{minipage} to prevent page or column breaking between or inside them. This should work fine as long you don't have footnotes in the paragraphs which would be placed at the bottom of the minipage.

format: plain


text:
# An Infinite Problem

Calculus Level 3

What is the value of $$\dfrac1{\infty}$$?

Extended real number is not used

×


text:
# Distance across the channel

Alignments to Content Standards: 8.F.B.4

In science class, Mrs. Winkler’s students are studying the water flow in a channel close to their school (see the photo below found here).

A cross section of the channel approximates the shape of an isosceles trapezoid with dimensions noted on the diagrams below: one showing an empty channel and the other the channel with water to the top of its banks.

The students found that there is a linear relationship between the water depth and the distance across the channel at water level. This means that we can write $y$, the distance across the channel at water level (in feet), as a function of the water depth, $d$ (in feet).

1. Use the information in the diagram to find two depth-distance pairs $(d, y)$ that correspond with each other. In other words, find two input-output pairs for the function.
2. Find an equation that describes $y$ as a function of $d$. Draw the graph of the equation.
3. Explain the meaning of the slope and the vertical intercept of the line in the context of the situation. In other words, what does each tell you about the depth of the water and the distance across the channel at water level?
4. What is the distance across the channel at water level when the water is 2 feet deep?

## IM Commentary

This task asks students to find a linear function that models something in the real world. After finding the equation of the linear relationship between the depth of the water and the distance across the channel, students have to verbalize the meaning of the slope and intercept of the line in the context of this situation. Students find this quite hard the first few times they are asked to interpret slopes and intercepts. It is important to give the students repeated opportunities to make connections between mathematical objects and their real world meaning. The slope gives us very specific information about how the depth of the water and the distance across the channel are related and we can describe it in very precise language: "For every one foot increase in water depth, the channel widens by 0.8 feet at water level." It is not enough to say "As the water gets deeper the channel becomes wider".

The task is meant as an instructional task as the complexity of the context (especially the language) would be difficult in an assessment situation. If used for instruction, students should be asked to share their interpretations of the slope and intercept. This could lead to a very productive discussion on precision of language and connections between formulas, graphs and the context of the problem.

There are several ways in which this task could be extended. First, it is possible to prove, using similar triangles, that the relationship between $d$ and $y$ is linear if the cross-section of the channel is an isosceles trapezoid. Another direction entirely would be for an instructor to present the students with the diagram of the channel and ask them to find the amount of water that flows down the channel under different conditions. In this form the task would also illustrate SMP 4 - Model with Mathematics.

## Solutions

Solution: Finding the equation of line

1. We can see from the picture that when there is no water in the channel, i.e. when $d=0$, the distance across the channel is $y=15$ feet. Similarly, when the depth of the water is $d=7.5$ feet, i.e. when the water is at its highest level, the distance across the channel at water level is $y=21$ feet. This gives us two input-output pairs that have to be on the line: $(0,15)$ and $(7.5, 21).$
2. Using the two points on the line, we can compute the slope of the line: $$m=(21-15)/(7.5-0) = 0.8.$$ The vertical intercept is 15, given by the point $(0,15).$ So we have $$y=0.8d+15$$ for the equation of the line. The graph is shown below:

3. The slope of the line is $0.8$ horizontal feet per vertical feet. This means that for every one foot increase in water depth, the distance across the channel at water level increases by 0.8 feet. The vertical intercept is $y=15$, this means that when the water depth is 0, i.e. when there is no water in the channel, the distance across the channel is 15 feet.
4. When the water is 2 feet deep the distance across the channel is $0.8\cdot 2+15 = 16.6$ ft.
5. A natural question to ask would be how much water is flowing down the channel during a specific amount of time. The answer depends on the depth of the channel and the velocity of the water. An easier to answer question would be "If the water has the same depth of 2 feet for a 10 foot stretch of the channel, how much water is in that part of the channel"

To compute the amount of water, i.e. the volume, we need to find the blue shaded area in the diagram for $d=2$ and multiply it by 10. We know that the area of a trapezoid with parallel sides of $15$ feet and $16.6$ feet and height $2$ feet is $2\cdot (15+16.6)/2 = 31.6$ square feet. Therefore, the volume of water we are looking for is $31.6\cdot 10 = 316$ cubic feet. Note that we are assuming that the shape of the channel stays the same for the entire 10 foot stretch.

After looking at a few specific questions like this, we can come up with a formula for the water flow down the channel, assuming that the depth of the water and the water velocity stay the same. With these assumptions we can compute how much water is flowing through a particular cross section of the channel at a fixed velocity (e.g. discharge in cubic feet per seconds): $$(\text{area of the cross section of the channel})\cdot (\text{velocity of the water})\cdot (\text{time}).$$

The USGS has a lot of data available for the depth and discharge of many waterways in the US at http://waterdata.usgs.gov/nwis/rt. They also provide a detailed explanation on the methods they use to compute water discharge at http://pubs.usgs.gov/fs/2005/3131/FS2005-3131.pdf

Solution: Alternate for (b)

When there is no water in the channel, the distance across the channel is 15 feet. This tells us that the vertical intercept of the line we are looking for is 15. Furthermore, we can see from the diagram, that when the depth of the water is 7.5 feet, the distance across the channel is 21 feet. This means that the distance increased by 6 feet when the depth increased by 7.5 feet. So for each additional feet of water depth, the distance increased by $\frac{6}{7.5} = 0.8$ feet. Therefore, the equation of the distance as a function of water depth is $$y=15+0.8d.$$


text:
# Linebreak in Package algorithm2e

I'm using the algorithm2e package for pseudocode. I got some very long lines that needed to be wrapped. Is there a way to indent the following (breaked) lines and/or put a mark at the first line to signalize that there happened a linebreak (similar to solution for listings package)

I found a similar question for the listings package, that was answered before, but I'm looking explicitly for an algorithm2e solution.

-
See also the related question if you want to skip line numbers for manually wrapped lines. – M.S. Dousti Jan 14 '14 at 22:04

The following minimal example defines

• \nosemic: remove printing of the end-of-line (EOL) character (typically ;);
• \dosemic: reinstate printing of the EOL character;
• \pushline: indents line by 1em (the typical indent); provided by algorithm2e as \Indp;
• \popline: remove indent of \pushline (undent?); provided by algorithm2e as \Indm

\documentclass{article}
\usepackage{algorithm2e}% http://ctan.org/pkg/algorithm2e
\makeatletter
\newcommand{\pushline}{\Indp}% Indent
\newcommand{\popline}{\Indm\dosemic}% Undent
\makeatother
\begin{document}
\begin{algorithm}[H]
\SetAlgoLined
\KwData{this text}
\KwResult{how to write algorithm with \LaTeX2e }
initialization\;
\While{not at end of this document}{
\eIf{understand}{
go to next section\;
\nosemic this is a very long statement that has to be\;
\pushline\dosemic wrapped over two lines\;
\popline current section becomes this one\;
}{
go back to the beginning of current section\;
}
}
\caption{How to write algorithms}
\end{algorithm}
\end{document}

Since I am unfamiliar with your algorithm2e setup, I'm not sure how you would like to combine the four commands. As such, I've kept them separate for maximum (yet manual) usage.

-

In fact there is a simpler solution: use classical \hangafter and \hangindent. This works fine with algorithm2e.

Notes:

1. \skiptext is the length of text indent in algorithm for a new block; so I choose 1/2 of this indent;

2. no need to redefine internal macro as \algocf@endline as almost all can be redefined in algorithm2e. Use \SetEndCharOfAlgoLine to redefine end of line

See below for an example of use of \hangafter and \hangindent:

\documentclass{article}
\usepackage{algorithm2e}% http://ctan.org/pkg/algorithm2e
%
\newcommand{\nosemic}{\SetEndCharOfAlgoLine{\relax}}% Drop semi-colon ;
\newcommand{\dosemic}{\SetEndCharOfAlgoLine{\string;}}% Reinstate
\newcommand{\pushline}{\Indp}% Indent
\newcommand{\popline}{\Indm\dosemic}% Undent
%
\begin{document}
\begin{algorithm}[H]
\SetAlgoLined
\KwData{this text}
\KwResult{how to write algorithm with \LaTeX2e }
initialization\;
\While{not at end of this document}{
\eIf{understand}{
go to next section\;
\nosemic this is a very long statement that has to be\;
\pushline\dosemic wrapped over two lines\;
\popline current section becomes this one\;
\hangindent=.5\skiptext\hangafter=1
this is a very long statement that should be indented automatically on the next
lines if this statement is longer than one line\;
Following lines should be indented normally\;
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go back to the beginning of current section\;
}
}
\caption{How to write algorithms}
\end{algorithm}
\end{document}

-
Welcome to TeX - LaTeX! – Werner Oct 11 '15 at 6:26

\Indp and \Indm can be used in the package algorithm2e for indent and removing indent. I have tested it.

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text:
# Special sequence

What if
2×9=1
3×9=2
4×9=3
5×9=4
6×9=5
7×9=6
8×9=7
9×9=8
10×9=9
then 1×9=???

x =  0

### Step-by-step explanation:

$x=0=0$

Did you find an error or inaccuracy? Feel free to write us. Thank you!

## Related math problems and questions:

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76% is between which of the following two numbers. 1) 1/5 & 1/4. 2) 1/3 & 1/2 3) 2/3 & 7/10 4) 3/4 & 5/6
• Median or middle
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• Sequence 11
What is the nth term of this sequence 1, 1/2, 1/3, 1/4, 1/5?
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How many 7-digit telephone numbers can be compiled from the digits 0,1,2,..,8,9 that no digit is repeated?
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Construct ABCD rhombus if its diagonal AC=9 cm and side AB = 6 cm. Inscribe a circle in it touching all sides...
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• Parenthesis
Fill in the examples correctly brackets: A) 7.9+12:3-2 =23 B) 7.9+12:3-2=75


text:
The inscribed angle θ is half of the central angle 2θ that subtends the same arc on the circle. The angle θ does not change as its vertex is moved around on the circle.

In geometry, an inscribed angle is the angle formed in the interior of a circle when two chords intersect on the circle. It can also be defined as the angle subtended at a point on the circle by two given points on the circle.

Equivalently, an inscribed angle is defined by two chords of the circle sharing an endpoint.

The inscribed angle theorem relates the measure of an inscribed angle to that of the central angle subtending the same arc.

The inscribed angle theorem appears as Proposition 20 on Book 3 of Euclid's Elements.

## Theorem

### Statement

For fixed points A and B, the set of points M in the plane for which the angle AMB is equal to α is an arc of a circle. The measure of ∠ AOB, where O is the center of the circle, is 2α.

The inscribed angle theorem states that an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

### Proof

#### Inscribed angles where one chord is a diameter

Case: One chord is a diameter

Let O be the center of a circle, as in the diagram at right. Choose two points on the circle, and call them V and A. Draw line VO and extended past O so that it intersects the circle at point B which is diametrically opposite the point V. Draw an angle whose vertex is point V and whose sides pass through points A and B.

Draw line OA. Angle BOA is a central angle; call it θ. Lines OV and OA are both radii of the circle, so they have equal lengths. Therefore, triangle VOA is isosceles, so angle BVA (the inscribed angle) and angle VAO are equal; let each of them be denoted as ψ.

Angles BOA and AOV add up to 180°, since line VB passing through O is a straight line. Therefore, angle AOV measures 180° − θ.

It is known that the three angles of a triangle add up to 180°, and the three angles of triangle VOA are:

180° − θ
ψ
ψ.

Therefore,

${\displaystyle 2\psi +180^{\circ }-\theta =180^{\circ }.}$

Subtract

${\displaystyle (180^{\circ }-\theta )}$

from both sides,

${\displaystyle 2\psi =\theta ,}$

where θ is the central angle subtending arc AB and ψ is the inscribed angle subtending arc AB.

#### Inscribed angles with the center of the circle in their interior

Case: Center interior to angle

Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.

Suppose this arc includes point E within it. Point E is diametrically opposite to point V. Angles DVE and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore,

${\displaystyle \angle DVC=\angle DVE+\angle EVC.}$

then let

${\displaystyle \psi _{0}=\angle DVC,}$
${\displaystyle \psi _{1}=\angle DVE,}$
${\displaystyle \psi _{2}=\angle EVC,}$

so that

${\displaystyle \psi _{0}=\psi _{1}+\psi _{2}.\qquad \qquad (1)}$

Draw lines OC and OD. Angle DOC is a central angle, but so are angles DOE and EOC, and

${\displaystyle \angle DOC=\angle DOE+\angle EOC.}$

Let

${\displaystyle \theta _{0}=\angle DOC,}$
${\displaystyle \theta _{1}=\angle DOE,}$
${\displaystyle \theta _{2}=\angle EOC,}$

so that

${\displaystyle \theta _{0}=\theta _{1}+\theta _{2}.\qquad \qquad (2)}$

From Part One we know that ${\displaystyle \theta _{1}=2\psi _{1))$ and that ${\displaystyle \theta _{2}=2\psi _{2))$. Combining these results with equation (2) yields

${\displaystyle \theta _{0}=2\psi _{1}+2\psi _{2}=2(\psi _{1}+\psi _{2})}$

therefore, by equation (1),

${\displaystyle \theta _{0}=2\psi _{0}.}$

#### Inscribed angles with the center of the circle in their exterior

Case: Center exterior to angle

The previous case can be extended to cover the case where the measure of the inscribed angle is the difference between two inscribed angles as discussed in the first part of this proof.

Given a circle whose center is point O, choose three points V, C, and D on the circle. Draw lines VC and VD: angle DVC is an inscribed angle. Now draw line VO and extend it past point O so that it intersects the circle at point E. Angle DVC subtends arc DC on the circle.

Suppose this arc does not include point E within it. Point E is diametrically opposite to point V. Angles EVD and EVC are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore,

${\displaystyle \angle DVC=\angle EVC-\angle EVD}$.

then let

${\displaystyle \psi _{0}=\angle DVC,}$
${\displaystyle \psi _{1}=\angle EVD,}$
${\displaystyle \psi _{2}=\angle EVC,}$

so that

${\displaystyle \psi _{0}=\psi _{2}-\psi _{1}.\qquad \qquad (3)}$

Draw lines OC and OD. Angle DOC is a central angle, but so are angles EOD and EOC, and

${\displaystyle \angle DOC=\angle EOC-\angle EOD.}$

Let

${\displaystyle \theta _{0}=\angle DOC,}$
${\displaystyle \theta _{1}=\angle EOD,}$
${\displaystyle \theta _{2}=\angle EOC,}$

so that

${\displaystyle \theta _{0}=\theta _{2}-\theta _{1}.\qquad \qquad (4)}$

From Part One we know that ${\displaystyle \theta _{1}=2\psi _{1))$ and that ${\displaystyle \theta _{2}=2\psi _{2))$. Combining these results with equation (4) yields

${\displaystyle \theta _{0}=2\psi _{2}-2\psi _{1))$

therefore, by equation (3),

${\displaystyle \theta _{0}=2\psi _{0}.}$

Animated gif of proof of the inscribed angle theorem. The large triangle that is inscribed in the circle gets subdivided into three smaller triangles, all of which are isosceles because their upper two sides are radii of the circle. Inside each isosceles triangle the pair of base angles are equal to each other, and are half of 180° minus the apex angle at the circle's center. Adding up these isosceles base angles yields the theorem, namely that the inscribed angle, ${\displaystyle \psi }$, is half the central angle, ${\displaystyle \theta }$.

### Corollary

By a similar argument, the angle between a chord and the tangent line at one of its intersection points equals half of the central angle subtended by the chord. See also Tangent lines to circles.

## Applications

The inscribed angle theorem is used in many proofs of elementary Euclidean geometry of the plane. A special case of the theorem is Thales' theorem, which states that the angle subtended by a diameter is always 90°, i.e., a right angle. As a consequence of the theorem, opposite angles of cyclic quadrilaterals sum to 180°; conversely, any quadrilateral for which this is true can be inscribed in a circle. As another example, the inscribed angle theorem is the basis for several theorems related to the power of a point with respect to a circle. Further, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal.

## Inscribed angle theorems for ellipses, hyperbolas and parabolas

Inscribed angle theorems exist for ellipses, hyperbolas and parabolas, too. The essential differences are the measurements of an angle. (An angle is considered a pair of intersecting lines.)

## References

• Ogilvy, C. S. (1990). Excursions in Geometry. Dover. pp. 17–23. ISBN 0-486-26530-7.
• Gellert W, Küstner H, Hellwich M, Kästner H (1977). The VNR Concise Encyclopedia of Mathematics. New York: Van Nostrand Reinhold. p. 172. ISBN 0-442-22646-2.
• Moise, Edwin E. (1974). Elementary Geometry from an Advanced Standpoint (2nd ed.). Reading: Addison-Wesley. pp. 192–197. ISBN 0-201-04793-4.


text:
# [texhax] No title for \listoffigures

Philip G. Ratcliffe philipratcliffe at tiscali.it
Fri Jun 27 12:54:32 CEST 2003

> Yes. I want to make my own section for listoffigures so I wrote:
>
> \renewcommand{\listfigurename}{}
> \section*{\mdseries \sffamily \bfseries Abbildungsverzeichnis}
> \listoffigures
> \newpage

You don't mention the packages you are using.  It is possible that one of
them redefines \listfigurename at the beginning of the document and so if
your \renewcommand{\listfigurename}{} is in the preamble it will get
overwritten.  Try putting the command just AFTER \begin{document}.  You
might also try the "tocbibind" package to avoid defining your own section
and so just have

\renewcommand\listfigurename{Abbildungsverzeichnis}

Cheers,  Phil Ratcliffe


text:
# high pass filter formula

Though you may choose to think of them as discrete elements, noise is a part of every waveform. High pass filters using LC components, i.e. So if we input an AC signal into the circuit from the function generator and place the signal This calculator can be used to design either low-pass filters or high-pass filters. {\displaystyle \Delta _{T}} , The expression for parameter 10KΩ Resistor. offer different resistances to signals input into them of different frequencies. A high-pass filter is usually modeled as a linear time-invariant system. High-Pass Filtering (Sharpening) A high-pass filter can be used to make an image appear sharper. It is sometimes called a low-cut filter or bass-cut filter in the context of audio engineering. Where f = operating frequency. high frequency signals pass through unimpeded without any attenuation. The formula is C = I / 2f Vpp. Δ The resultant images by BHPF is much sharper than GHPF ,while analysis the FFT of CT and MRI image, one sharp spike is concentrated in the middle. = The DC is only needed to power the microphone on and should not appear Definition High Pass Filter. Formula – band pass filter … It passes through high frequency signals. R where fc is in hertz, τ is in seconds, R is in ohms, and C is in farads. Butterworth filters. . How to Calculate the Voltage Across a Capacitor, How to Calculate the Current Through a Capacitor, How to Calculate the Voltage Across a Capacitor. 1 . The transition between the pass-band and stop-band of a first order filter with cut-off frequency is characterized by the the slope of 20 dB per decade of frequency change. 1 In power supply circuits, this capacitor can be calculated to ensure the least ripple at the output. A high pass filter composed α α This type of filter has a transfer function of the first order. be represented by the sequence The difference between Butterworth and Gaussian filters is that the former is much sharper than latter. inductive reactance is how the above circuit operates. 1 High Pass Filter Formula. Low & High Pass Filter. − This electronics video tutorial discusses how resistors, capacitors, and inductors can be used to filter out signals according to their frequency. By arranging components in this way, high-frequency signals can pass, while the capacitor blocks any frequencies that are too low. Microphones need DC power in order to have the power to turn on and be able to operate. 1 The capacitor’s impedance (Figure above) increases with decreasing frequency. Discrete-time filter design is beyond the scope of this article; however, a simple example comes from the conversion of the continuous-time high-pass filter above to a discrete-time realization. R Thus also takes advantage of the fact that the DFT of a Gaussian function is also a Gaussian function shown in figure 6,7,8,9. the inductor. low frequencies. Op amp high pass filters are easy to implement using just a few electronic components being used in a variety of electronic devices and circuits to eliminate hum and other noise. {\displaystyle t} Display the original and filtered signals, and also their spectra. As with low-pass filters, high-pass filters have a rated cutoff frequency, above which the output voltage increases above 70.7% of the input voltage. The basic formula for calculating an RC high pass is: $$\frac{V_{out}}{V_{in}} = \frac{R}{Z}$$ The following applies to the impedance Z: $$Z = \sqrt{R^2 + X_C^2}$$ The RC high pass filter transfer function is calculated according to: What is it, Circuit, formulas, curve? From an equivalent network point of view, the design of a high-pass network is quite straightforward as it is sufficient to interchange the topological position of inductors and capacitors of the low-pass filter. You don't need to use Matlab, Octave or even VBA and for such an undemanding application you almost certainly don't need to learn the intricacies of FIR filter design. High pass filter give emphasis on the high frequencies in the image. c The above circuit uses two first-order filters connected or cascaded together to form a second-order or two-pole high pass network. , Just as in the case of the capacitive low-pass filter circuit, the capacitive high-pass filter’s cutoff frequency can be found with the same formula: To build a high pass RL filter, the inductor is placed in parallel to the power source signals entering the circuit, as shown below For simplicity, assume that samples of the input and output are taken at evenly spaced points in time separated by It passes through high frequencies but block Low Pass Filter Calculator 0.5 6 One of such filters is High Pass Filter. As an example, the formula above, applied to a tweeter with R=10 Ohm, will determine the capacitor value for a cut-off frequency of 5 kHz. This can be achieved using the formula: X c = 1 2 π f C α 25% Off on Electrical Engineering Shirts. Since the formula for a high-pass RL filter is f= R/2πL, doing the math we get, f= R/2πL= 10KΩ/ (2 (3.14) (470mH))= 3,388 Hz, which is approximately 3.39KHz. , then The way a high-pass / low cut works is it rolls off (removes) all the frequencies below a certain point. {\displaystyle V_{\text{out}}} Since the formula for a high-pass RL filter is f= R/2πL, doing the math we get, f= R/2πL= 10KΩ/(2(3.14)(470mH))= 3,388 Hz, which is approximately 3.39KHz. Although any form of filter response can be chosen, the Butterworth response simplifies the equations and the electronic design process producing the electronic component values can be undertaken in a … To achieve better selectivity, we can cascade a set of such first order filters to form an nth order filter with a slope of 20n dB per decade. To build a high pass filter, the components we will use are a function generator, a 10nF ceramic capacitor, and a . There are many applications for an RLC circuit, including band-pass filters, band-reject filters, and low-/high-pass filters. And low resistance to high-frequency signals. Band-Pass & Reject Filter Equation and Formulas. This means all frequencies above 3.39KHz will be passed through without attentuation, while frequencies below this value will begin to get attenuated. The cutoff frequency is where the pole of the filter levels off the filter's frequency response. Frequencies below 15,923Hz will be attenuated. C When such a filter is built into a loudspeaker cabinet it is normally a passive filter that also includes a low-pass filter for the woofer and so often employs both a capacitor and inductor (although very simple high-pass filters for tweeters can consist of a series capacitor and nothing else). {\displaystyle \alpha =0.5} This type of capacitor also functions as a coupling capacitor × to check the signal. High-pass filters have many applications. Thus, low-frequency signals pass through very easily without any attenuation and high frequency signals are either , and let The filter recurrence relation provides a way to determine the output samples in terms of the input samples and the preceding output. Therefore, low-frequency α This is only possible if you use only one energy storage element like inductor and capacitor. getting through, which is why it's a high pass filter. High Pass filter allows the frequencies which are higher than the cut off frequency ‘fc’ and blocks the lower frequency signals. Some of the circuits have electrical filters which perform signal processing by emphasizing signals in certain ranges and rejecting signals in other frequency ranges out of a mix of different frequencies. This post presents C# code for a fourth order zero-phase shift low-pass Butterworth filter function. y = highpass (x,wpass) filters the input signal x using a highpass filter with normalized passband frequency wpass in units of π rad/sample. ( Introduction. As with low-pass filters, high-pass filters have a rated cutoff frequency, above which the output voltage increases above 70.7% of the input voltage. ≤ of a resistor and a capacitor is called a high pass RC filter. The following formula is used to calculate the cutoff frequency of a high pass filter. They are used as part of an audio crossover to direct high frequencies to a tweeter while attenuating bass signals which could interfere with, or damage, the speaker. To use this calculator, all a user must do is enter in any 2 values, and the calculator will compute the third field. High pass filters are very common and are used in many type of circuit setups. A low-pass filter is a circuit that allows low-frequency signals and stops high-frequency signals. Thus, they deal with both type of signals. (Figure below) This high impedance in series tends to block low-frequency signals from getting to load. The cut-off frequency, corner frequency or -3dB point of a high pass filter can be found using the standard formula of: ƒc = 1/(2πRC). The high frequency signals go through unimpeded and pass to output. Specify the desired filter type (low-pass, high-pass, or band-pass) in the constructor, along with the other needed parameters: the number of taps, the transition frequencies, and the sampling frequency of the data you’ll be filtering. while filtering low-frequency signals through because it couples the AC signal from one part of a circuit to another, while blocking the DC. So, as the name suggests, it is a filter that will block Low frequencies, but pass the high frequency above the predetermined value, which will be calculated by the formula.. One of such filters is High Pass Filter. 3.18 Up tp 93% Off - Launching Official Electrical Technology Store - Shop Now! First, we will take a look at the phase response of the transfer equations. fc = 1 / (2*pi*R*C) Where fc is the cutoff frequency (hz) Therefore, this RC circuit will pass frequencies above 15,923Hz with barely no attenuation. These filters emphasize fine details in the image – exactly the opposite of the low-pass filter. ", ECE 209: Review of Circuits as LTI Systems, https://en.wikipedia.org/w/index.php?title=High-pass_filter&oldid=996461960, Creative Commons Attribution-ShareAlike License, A large α implies that the output will decay very slowly but will also be strongly influenced by even small changes in input. A low-pass filter (LPF) is a filter that passes signals with a frequency lower than a selected cutoff frequency and attenuates signals with frequencies higher than the cutoff frequency. DC signals, it blocks them from entering through, as you can see in the circuit diagram above. , Inductors t Same as like before, if you look into the name it shows “Passive”, “High”, “Pass” and “Filter”. A high-pass filter (HPF) is an electronic filter that passes signals with a frequency higher than a certain cutoff frequency and attenuates signals with frequencies lower than the cutoff frequency. Since inductors offer such high resistance to high frequency signals, current signals of high frequency will not go through the inductor of this circuit. C An RC high-pass filter, also known as an RC Differentiator, works oppositely. [8] The unsharp masking, or sharpening, operation used in image editing software is a high-boost filter, a generalization of high-pass. and not DC. This is the schematic of the circuit we will build, shown below: So the formula to calculate the frequency of an RC circuit is, frequency= 1/2πRC. to a very low frequency signal such as 100Hz or so, the capacitor will block out this voltage signal and it will not go through to output or be {\displaystyle V_{\text{in}}} Some of the circuits have electrical filters which perform signal processing by emphasizing signals in certain ranges and rejecting signals in other frequency ranges out of a mix of different frequencies. High-Pass Filtering (Sharpening) A high-pass filter can be used to make an image appear sharper. A high pass filter is a filter which passes high-frequency signals and blocks, or impedes, low-frequency Formula – RC high pass filter calculation. As you get lower and lower and further away from 3.39KHz, {\displaystyle Q_{c}(t)} By definition, Q which correspond to the same points in time. Because this filter is active, it may have non-unity passband gain. , Substituting Equation (Q) into Equation (I) and then Equation (I) into Equation (V) gives: This equation can be discretized. Main writes that DI unit inputs (as opposed to microphone inputs) do not need high-pass filtering as they are not subject to modulation by low-frequency stage wash—low frequency sounds coming from the subwoofers or the public address system and wrapping around to the stage. Cutoff Frequency of a Low Pass Filter. As a capacitor is a reactive device, it offers differing resistance to signals of different frequencies entering through it. lowpass(x,150,fs) Lowpass Filtering of Musical Signal. 2 Definition High Pass Filter. Here's the corresponding high-pass filter: xfilt = filter([1-a a-1],[1 a-1], x); If you need to design a filter, and have a license for the Signal Processing Toolbox, there's a bunch of functions, look at fvtool and fdatool. and , then the And this is why this circuit is a high-pass filter circuit. If we add an Amplifier across passive high pass filter, we can easily create Active high pass filter. From an equivalent network point of view, the design of a high-pass network is quite straightforward as it is sufficient to interchange the topological position of inductors and capacitors of the low-pass filter. Such low-current and low-voltage line level crossovers are called active crossovers. When such a filter is built into a loudspeaker cabinet it is normally a passive filter that also includes a low-pass filter for the woofer and so often employs both a capacitor and inductor (although very simple high-pass filters for tweeters can consist of a series capacitor and nothing else). R Below you can see a video of an RC high pass filter. High Pass Filter plays an important role in achieving best sound quality. output with full gain, no attenuation, while greatly attenuating low-frequency signals. An alternative, which provides good quality sound without inductors (which are prone to parasitic coupling, are expensive, and may have significant internal resistance) is to employ bi-amplification with active RC filters or active digital filters with separate power amplifiers for each loudspeaker. y High and low pass filters are simply connected in series. Please note that, in this article, we will use "stop-band attenuation" and "the minimum stop-band attenuation" interchangeably. y At very high frequencies the capacitor acts as a short circuit and all the input appears at the output. As it offers very high resistance to And they record The Band Pass Filter – the band pass filter allows signals falling within a certain frequency band setup between two points to pass through while blocking both the lower and higher … With the default signals (400 Hz, 150 Hz, and 30 Hz) you should be able to filter each one out successfully. Capacitor Impedance Calculator ≈ If you now The transfer function of this linear time-invariant system is: In this case, the filter has a passband gain of -R2/R1 and has a cutoff frequency of. If you leave the filter coefficients unchanged and you change the sampling frequency, also the cut-off frequency will change. The band pass filter passes a band of frequencies between a lower cutoff frequency, f l, and an upper cutoff frequency, f h. Frequencies below f l and above f h are in the stop band. The output voltage $$V_{out}$$ is tapped behind both filters. ≪ ) In other words, high-frequency signals go through much easier and low-frequency signals have a much harder High-pass and low-pass filters are also used in digital image processing to perform image modifications, enhancements, noise reduction, etc., using designs done in either the spatial domain or the frequency domain. x ( High-pass filters are complementary to low-pass filters. One amplifier, the professional audio model DC300 made by Crown International beginning in the 1960s, did not have high-pass filtering at all, and could be used to amplify the DC signal of a common 9-volt battery at the input to supply 18 volts DC in an emergency for mixing console power. 10 The phase angle of the resulting output signal at ƒc is +45 o. fc = 1 / (2*pi*R*C) Where fc is the cutoff frequency (hz) A high pass filter attenuates the low-frequency signal and allows only high-frequency signal to pass through it. 5000 High-pass filters would have similar phase responses, shifted by 180°. = 10 In power supply circuits, this capacitor can be calculated to ensure the least ripple at the output. Active High Pass Filter – 1st Order & 2nd Order Active High Pass Filters. Single pole high pass op amp filters The cut off frequency or break point of the filter can be calculated very easily by working out the frequency at which the reactance of the capacitor equals the resistance of the resistor. A max = pass band gain of the filter = 1 + (R 3 /R 2). For example, noises (e.g., footsteps, or motor noises from record players and tape decks) may be removed because they are undesired or may overload the RIAA equalization circuit of the preamp. These high pass filters have many applications at high frequency ranges of 10 MHz. below: The above is a high pass filter. In this circuit, instead of the high-frequency signals going through the inductor and {\displaystyle C={\frac {1}{2\pi fR}}={\frac {1}{6.28\times 5000\times 10}}=3.18\times 10^{-6}} For high pass filter, this circuit is similar to low-pass filter circuit, but the position for resistors and capacitor are interchanged. A low frequency is where the pole of the filter up during recording band! The response of the resulting output signal at ƒc is +45 o, frequencies. Practical example of building one articles will examine phase relationships in band-, notch- ( ). Of a resistor and practically reactance of the capacitor acts as a short circuit and all input. C Definition high pass RL filter is, let 's go over a practical example of building one,. Differentiator, works oppositely fine details in the image – exactly the same order and! Function will be passed through without attentuation, while the capacitor will have no effect and the! After the 2 values are entered in, the waveshaping spectra of Section 5.3 always. Can check this if you leave the filter 's frequency response components are relatively isolated to a specific frequency.. Filtered signals, such as blocking DC from circuitry sensitive to non-zero average voltages or radio frequency.... Be discretized capacitor or inductor: x C = I / 2f Vpp relation provides a way to determine output... 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text:
# Capitalise letter when using cref (cleveref) for subfig

Is it possible to have cleveref capitalise the letter when cross referencing a subfigure?

Currently my subfigures are configured to capitalise the letter in a subfig (i.e A B C D E) but when i use the \cref command it will generated Fig. 1a.

Here is MWE:

\documentclass[a4paper]{article}
\usepackage{subcaption}
\usepackage{caption}
\captionsetup{justification=centerlast,font=small,labelfont=sc,margin=50pt}
\usepackage[capitalise]{cleveref}

\begin{document}

\begin{figure}
\centering
\subcaptionbox{\label{subfig:Test1}}[0.49\textwdith]{\includegraphics[width=0.49\textwidth]{Fig.png}}
\subcaptionbox{}[0.49\textwdith]{\includegraphics[width=0.49\textwidth]{Fig2.png}}

I would like this cross reference to show a capital 'A' rather than a lower case: \cref{subfig:Test1}.

\end{document

• Can you please add a minimal example of what you're doing up to now? Dec 18 '19 at 16:56
• @egreg I have added the MWE Dec 19 '19 at 10:16

The problem is that you are not labelling the subfigures with a capital letter, but labelling with a small letter and then printing it in the caption with the small caps font. If you wish to use the small caps font for this label then rather than (or in addition to) setting the labelfont you should redefine the printing of the subfigure label via

\renewcommand{\thesubfigure}{\scshape\alpha{subfigure}}

so this formatting is used elsewhere when you reference the label.

\documentclass[a4paper]{article}

\usepackage{graphicx}
\usepackage{subcaption}
\usepackage{caption}
\captionsetup{justification=centerlast,font=small,margin=50pt}
\renewcommand{\thesubfigure}{\scshape\alph{subfigure}}
\usepackage[capitalise]{cleveref}

\begin{document}

\begin{figure}
\centering
\subcaptionbox{\label{subfig:Test1}}[0.49\textwidth]{\includegraphics[width=0.49\textwidth]{example-image-a}}
\subcaptionbox{}[0.49\textwidth]{\includegraphics[width=0.49\textwidth]{example-image-b}}
\end{figure}

I would like this cross reference to show a capital 'A' rather than a
lower case: \cref{subfig:Test1}.

\end{document}

On the other hand if you really want to label by capital letters, then the appropriate redefinition is

\renewcommand{\thesubfigure}{\Alph{subfigure}}

regardless of the caption font.

• Yes of course. I see now. thanks for the help! Dec 19 '19 at 12:14


text:
# Surface Area of Simple 3D Shapes07:22 minutes

Video Transcript

## Videos in this Topic

Use Equations and Inequalities to Solve Geometry Problems (9 Videos)

## Surface Area of Simple 3D Shapes Exercise

### Would you like to practice what you’ve just learned? Practice problems for this video Surface Area of Simple 3D Shapes help you practice and recap your knowledge.

• #### Find the surface area of each prism.

##### Hints

The area of the top and bottom of the rectangular prism is,

• $\text{Area}=2(16\cdot 12)~\text{in}^2$
• $\text{Area}=2(192)~\text{in}^2$
• $\text{Area}=384~\text{in}^2$

The perimeter, $P$, of an isosceles triangle with leg lengths $15~\text{in}$ and base $10~\text{in}$ is,

• $P=15~\text{in}+15~\text{in}+10~\text{in}$
• $P=40~\text{in}$

The area for the circular bases of the cylinder is,

• $A=2(\pi \cdot 3^2)~\text{in}^2$
• $A=18\pi ~\text{in}^2$
• $A\approx18(3.14)~\text{in}^2$
• $A\approx56.52~\text{in}^2$

##### Solution

Rectangular Prism

To find the area of the rectangular prism use the following steps:

1. Recall that the area of a rectangle is found by multiplying the length by the width.

2. Find the area of the following faces:

• Let length $l=22$, width $w=19$, and height $h=6$.
• Top: $A=lw=22~\text{in}\cdot 19~\text{in}=418~\text{in}^2$
• The areas for the top and bottom of the rectangular prism are the same. So, we can just multiply the area of the top by $2$ to get the area of both the top and bottom. Now we can just find the area of the six faces.
• Top & Bottom: $A=2(lw)=2(22~\text{in}\cdot 19~\text{in})=2(418)~\text{in}^2=836~\text{in}^2$
• Front & Back: $A=2(wh)=2(19~\text{in}\cdot 6~\text{in})=2(114)~\text{in}^2=228~\text{in}^2$
• Left & Right:$A=2(lh)=2(22~\text{in}\cdot 6~\text{in})=2(132)~\text{in}^2=264~\text{in}^2$
3. Add all of the areas together:
• $\text{Surface Area}=836~\text{in}^2+228~\text{in}^2+264~\text{in}^2$
• $\text{Surface Area}=1,328~\text{in}^2$
Therefore, we can use the formula $\text{Surface Area}=2lw+2wh+2lh$ to represent surface area of a rectangular prism.

$~$

Triangular Prism

To find the area of the triangular prism use the following steps:

1. Recall that the area of a triangle is found by multiplying the base and height by $\frac{1}{2}$.

2. Find the area of the front and back triangles:

• Use the area of a triangle formula, $\frac{1}{2}bh$, and multiply it by $2$ to represent both the front and back.
• Let base $b=23$, height $h=29$.
• Front & Back:$A=2(\frac{1}{2}bh)=2(\frac{1}{2}\cdot 23~\text{in}\cdot 29~\text{in})=2(333.5)~\text{in}^2=667~\text{in}^2$
3. Find the area of the lateral face by unfolding the sides to create a long rectangle. The height of this new rectangle is $2~\text{in}$ and the length is found by finding the perimeter, $P$, of all three sides of the isosceles triangle.
• Lateral Face: $A=hP=2~\text{in}(31~\text{in}+31~\text{in}+23~\text{in})=170~\text{in}^2$
• $\text{Surface Area}=667~\text{in}^2+170~\text{in}^2=837~\text{in}^2$
Therefore, we can use the formula $\text{Surface Area}=2A_{\triangle}+hP$ to represent surface area of a triangular prism.

$~$

Cylinder

To find the area of the cylinder, use the following steps:

1. Recall that the area of a circle is found by using the formula $A=\pi r^2$.

2. Find the area of the two circular bases:

• Use the area of a circle formula and multiply it by $2$ to represent both bases.
• Bases: $A=2(\pi \cdot 4^2)~\text{in}^2=2(16\pi)~\text{in}^2=32\pi~\text{in}^2$
3. Find the area of the lateral face by unrolling the face of the cylinder. This creates a rectangle with a height of $16~\text{in}$ and a length that is the same as the circumference, $C$, of the circular base. Recall that $C=2\pi r$.
• Base:$C=2\pi \cdot 4~\text{in}=8\pi ~\text{in}$
• Lateral Face: $A=16~\text{in}(8\pi ~\text{in})=128\pi~\text{in}^2$
4. Add both areas together and use $3.14$ to approximate $\pi$:
• $\text{Surface Area}\approx 32(3.14)~\text{in}^2+128(3.14)~\text{in}^2\approx 100.48~\text{in}^2+401.92~\text{in}^2=502.4~\text{in}^2$
Therefore, we can use the formula $\text{Surface Area}=2A_{\circ}+hC$ to represent surface area of a cylinder.

• #### Identify the formula for the surface area of each prism.

##### Hints

The surface area for all prisms can be found by doubling the area of the base and adding the lateral area.

The area of the base can be found by identifying the shape and using the area formula for that particular shape.

The base of a cylinder is a circle.

##### Solution

The surface area for all prisms can be found by doubling the area of the base, the shape that clarifies the 3D object, and adding the lateral area.

This is represented by the formula, $SA=2A_{\text{base}}+Ph$ where $P$ is the perimeter of the lateral face and $h$ is the height of the lateral face.

1. The formula for the surface area of a cylinder is, $SA=2(\pi r^2)+Ph$.

• The base of a cylinder is a circle.
• The area of a circle is, $A=\pi r^2$.
• The lateral face is a rectangle represented by $Ph$ where $P$ is actually the circumference of the circular base, $C=2 \pi r$, and $h$ is the height of the cylinder.
2. The formula for the surface area of a rectangular prism is, $SA=2lw+2wh+2lh$.
• The base of a rectangular prism is a rectangle.
• The area of a rectangle is $A=lw$
• There are four lateral faces that are all rectangles, the front, back, left, and right of the prism which represent $2wh$ and $2lh$.
3. The formula for the surface area of a triangular prism is, $SA=2(\frac{1}{2}bh)+Ph$.
• The base of a triangular prism are two triangles.
• The area of a triangle is, $A=\frac{1}{2}bh$ where $b$ represents the base and $h$ represents the height.
• The lateral face is a rectangle formed from unfolding the sides and is represented by $Ph$ where $P$ is the perimeter of all $3$ sides and $h$ is the height of the unfolded side.
4. Another formula for the surface area of a cylinder is, $SA\approx2(3.14\cdot r^2)+Ph$.
• The base of a cylinder is a circle.
• The area of a circle is, $A=\pi r^2$.
• $3.14$ is used an approximation for $\pi$.
• The lateral face is a rectangle represented by $Ph$ where $P$ is the circumference of the circular base and $h$ is the height of the cylinder.

• #### Determine the surface area of each prism.

##### Hints

$3.14$ can be used as an approximation for pi, $\pi$.

The surface area of the cylinder is found by solving the following expression,

• $SA \approx 2(\pi\cdot 2^2)~\text{in}^2+(2\pi \cdot 2)(10)~\text{in}^2$

The surface area of the triangular prism is found by solving the following expression,

• $SA=2\left( \frac{1}{2}(16\cdot 15)~\text{in}^2 \right)+(16+17+17)(1)~\text{in}^2$

##### Solution

1. Cylinder with radius $7~\text{in}$ and height $17~\text{in}$.

• The surface area is found by using the formula, $SA=2\text{Area}_{\text{circle}}+Ph$.
• $Ph$ represents the lateral face of the cylinder found by unrolling the face of the cylinder into a rectangle.
• The rectangle is composed of the circumference, which we're using the perimeter, $P$, of the circular base, and the height $h$ of the cylinder.
• The area of a circle is $\pi r^2$ and the circumference of a circle is $2\pi r$.
• $SA=2(\pi \cdot 7^2)+(2\pi \cdot 7)\cdot 17$
• $SA \approx 2(3.14\cdot 49)+(2\cdot 3.14\cdot 7)\cdot 17$
• $SA \approx 307.72+747.32$
• $SA \approx 1,055.04~\text{in}^2$
2. Triangular Prism with base $18~\text{in}$, height $12~\text{in}$, legs $15~\text{in}$, and depth $3~\text{in}$.
• The surface area is found by using the formula, $SA=2\text{Area}_{\text{triangle}}+Ph$.
• $Ph$ represents the lateral faces of the prism found by placing 1 triangular face on ground and unfolding the sides into 1 long rectangle.
• The rectangle is composed of the perimeter $P$ of the isosceles triangle, and the depth $h$ of the triangular prism.
• $SA=2\cdot \frac{1}{2}(18\cdot 12)+(15+15+18)(3)$
• $SA=216+144$
• $SA=360~\text{in}^2$
3. Rectangular Prism length $21~\text{in}$, width $12~\text{in}$, and height $17~\text{in}$.
• The surface area is found by using the formula, $SA=2lw+2wh+2lh$, where $l$ represents the length, $w$ represents the width, and $h$ represents the height.
• The surface area is basically adding the areas of all 6 rectangular faces, front, back, top, bottom, left and right.
• $SA=2(21\cdot 12)+2(12\cdot 17)+2(21\cdot 17)$
• $SA=2(252)+2(204)+2(357)$
• $SA=504+408+714$
• $SA=1,626~\text{in}^2$
4. Cylinder with radius $5~\text{in}$ and height $10~\text{in}$.
• The surface area is found by using the formula, $SA=2\text{Area}_{\text{circle}}+ph$.
• $Ph$ represents the lateral face of the cylinder found by unrolling the face of the cylinder into a rectangle.
• The rectangle is composed of the circumference, which we use the perimeter, $P$, of the circular base, and the height $h$ of the cylinder.
• The area of a circle is $\pi r^2$ and the circumference of a circle is $2\pi r$.
• $SA=2(\pi \cdot 5^2)+(2\pi \cdot 5)\cdot 10$
• $SA \approx 2(3.14\cdot 25)+(2\cdot 3.14\cdot 5)\cdot 10$
• $SA=157+314$
• $SA=471~\text{in}^2$
5. Triangular Prism with base $10~\text{in}$, height $12~\text{in}$, legs $13~\text{in}$, and depth $2~\text{in}$.
• The surface area is found by using the formula, $SA=2\text{Area}_{\text{triangle}}+Ph$.
• $Ph$ represents the lateral faces of the prism found by unfolding the sides into 1 long rectangle.
• The rectangle is composed of the perimeter $P$ of the isosceles triangle, and the depth $h$ of the triangular prism.
• $SA=2\cdot \frac{1}{2}(10\cdot 12)+(13+13+10)(2)$
• $SA=120+72$
• $SA=192~\text{in}^2$

• #### Calculate the surface area of the given composite 3D object.

##### Hints

The surface area for all prisms can be found by doubling the area of the base and adding the lateral area.

Remember that the radius of a circle is half the diameter.

The total surface area is represented by four cylinders, one triangular prism, and one rectangular prism. Don't forget to subtract the area of any shared sides that are not on the surface.

##### Solution

To find the surface area of the 3D figure, we need to calculate the surface of the individual prisms and cylinders, then add them all together. There are 4 cylinders, 1 triangular prism, and 1 rectangular prism.

1. Find the surface area of the 4 cylinders with diameter $2~\text{in}$ and height $5~\text{in}$.

• The surface area is found by using the formula, $SA=2(\pi r^2)+Ph$.
• $Ph$ represents the lateral face of the cylinder found by unrolling the face of the cylinder into a rectangle.
• The rectangle is composed of the circumference $P$ of the circular base, and the height $h$ of the cylinder.
• The area of a circle is $\pi r^2$ and the circumference of a circle is $2\pi r$.
• Since the diameter is twice the radius, we know that $r=1~\text{in}$.
• $SA=2(\pi \cdot 1^2)~\text{in}^2+(2\pi \cdot 1)\cdot 5~\text{in}^2$
• $SA\approx 2(3.14\cdot 1)~\text{in}^2+(2\cdot 3.14\cdot 1)\cdot 5~\text{in}^2$
• $SA\approx 6.28~\text{in}^2+31.4~\text{in}^2$
• $SA\approx 37.68~\text{in}^2$
• The surface area of 1 cylinder is $37.68~\text{in}^2$ which makes the surface area of 4 cylinders, $4(37.68~\text{in}^2)=150.72~\text{in}^2$.
• Since the top and bottom of all four cylinders are being shared with the other objects and therefore not on the surface, subtract the area of $8$ circles from the total surface area.
• $\text{Area}=8\pi r^2=8(\pi \cdot (1~\text{in})^2)=8\pi~\text{in}^2\approx 8(3.14)~\text{in}^2\approx 25.12~\text{in}^2$.
• The new surface area is $150.72~\text{in}^2-25.12~\text{in}^2=125.60~\text{in}^2$.
2. Find the surface area of the triangular prism with base $10~\text{in}$, height $12~\text{in}$, legs $13~\text{in}$, and depth $3~\text{in}$.
• The surface area is found by using the formula, $SA=2(\frac{1}{2}bh_{1})+Ph_{2}$.
• $Ph_{2}$ represents the lateral faces of the prism found by placing 1 triangular face on ground and unfolding the sides into 1 long rectangle.
• The rectangle is composed of the perimeter $P$ of the isosceles triangle, and the depth $h_{2}$ of the triangular prism.
• $SA=2\cdot \frac{1}{2}(10\cdot 12)~\text{in}^2+(13+13+10)(3)~\text{in}^2$
• $SA=120~\text{in}^2+36(3)~\text{in}^2$
• $SA=120~\text{in}^2+108~\text{in}^2$
• $SA=228~\text{in}^2$
3. Find the surface area of the rectangular prism length $10~\text{in}$, width $3~\text{in}$, and height $4~\text{in}$.
• The surface area is found by using the formula, $SA=2lw+2wh+2lh$, where $l$ represents the length, $w$ represents the width, and $h$ represents the height.
• The surface area is basically adding the areas of all 6 rectangular faces, front, back, top, bottom, left and right.
• $SA=2(10\cdot 3)~\text{in}^2+2(3\cdot 4)~\text{in}^2+2(10\cdot 4)~\text{in}^2$
• $SA=2(30)~\text{in}^2+2(12)~\text{in}^2+2(40)~\text{in}^2$
• $SA=60~\text{in}^2+24~\text{in}^2+80~\text{in}^2$
• $SA=164~\text{in}^2$
4. Add the surface areas together to find the total surface area of the 3D object.
• $SA=125.60~\text{in}^2+228~\text{in}^2+164~\text{in}^2$
• $SA=517.60~\text{in}^2$

• #### Compute the area of each shape.

##### Hints

$3.14$ can be used as an approximation for pi, $\pi$.

The radius, $r$, is a straight line from the center to the circumference of a circle.

The area for a triangle with a base of $4~\text{in}$ and height of $8~\text{in}$ is $16~\text{in}^2$.

##### Solution

Green Circle

• The area formula for a circle is, $A=\pi r^2$, where $r$ represents the radius.
• The image shows that the radius is $3~\text{in}$.
• $\text{A}=\pi (3^2)~\text{in}^2$
• $\text{A}\approx 3.14(9)~\text{in}^2$
• $\text{A}\approx 28.26~\text{in}^2$
Orange Triangle
• The area formula for a triangle is, $A=\frac{1}{2}bh$, where $b$ represents the base and $h$ represents the height.
• The image shows that the base is $6~\text{in}$ and the height is $4~\text{in}$.
• $\text{A}=\frac{1}{2}(6)(4)~\text{in}^2$
• $\text{A}=3(4)~\text{in}^2$
• $\text{A}=12~\text{in}^2$
Blue Rectangle
• The area formula for a rectangle is, $A=lw$, where $l$ represents the length and $w$ represents the width.
• The image shows that the length is $10~\text{in}$ and the width is $6~\text{in}$.
• $\text{A}=10(6)~\text{in}^2$
• $\text{A}=60~\text{in}^2$

• #### Find the surface area of the composite 3D object.

##### Hints

The surface area for all prisms can be found by doubling the area of the base and adding the lateral area.

Unraveling the lateral faces of a prism normally creates a rectangular figure.

The total surface area of the 3D figure is represented by a rectangular prism and a triangular prism. Don't forget to subtract the area of any shared sides that are not on the surface.

##### Solution

To find the surface area of the 3D figure, we need to calculate the surface of the individual prisms, then add them all together, and subtracting the area of any sides that are not on the surface. There is 1 triangular prism and 1 rectangular prism.

1. Find the surface area of the rectangular prism length $8~\text{cm}$, width $10~\text{cm}$, and height $4~\text{cm}$.

• The surface area is found by using the formula, $SA=2lw+2wh+2lh$, where $l$ represents the length, $w$ represents the width, and $h$ represents the height.
• The surface area is basically adding the areas of all 6 rectangular faces, front, back, top, bottom, left and right.
• $SA=2(8\cdot 10)~\text{cm}^2+2(10\cdot 4)~\text{cm}^2+2(8\cdot 4)~\text{cm}^2$
• $SA=2(80)~\text{cm}^2+2(40)~\text{cm}^2+2(32)~\text{cm}^2$
• $SA=160~\text{cm}^2+80~\text{cm}^2+64~\text{cm}^2$
• $SA=304~\text{cm}^2$
• Since the top of the rectangular prism is a shared side and not on the surface, subtract the rectangular area from the total surface area.
• $\text{Area}=8~\text{cm}\cdot 10~\text{cm}=80~\text{cm}^2$
• The new surface area is $304~\text{cm}^2-80~\text{cm}^2=224~\text{cm}^2$.
2. Find the surface area of the triangular prism with base $8~\text{in}$, height $3~\text{in}$, legs $5~\text{in}$, and depth $10~\text{in}$.
• The surface area is found by using the formula, $SA=2(\frac{1}{2}bh_{1})+Ph_{2}$.
• $Ph_{2}$ represents the lateral faces of the prism found by placing 1 triangular face on ground and unfolding the sides into 1 long rectangle.
• The rectangle is composed of the perimeter $P$ of the isosceles triangle, and the depth $h_{2}$ of the triangular prism.
• $SA=2\cdot \frac{1}{2}(8\cdot 3)~\text{cm}^2+(5+5+8)(10)~\text{cm}^2$
• $SA=24~\text{cm}^2+18(10)~\text{cm}^2$
• $SA=24~\text{cm}^2+180~\text{cm}^2$
• $SA=204~\text{cm}^2$
• Since the bottom of the triangular prism is a shared side and not on the surface, subtract the rectangular area from the total surface area.
• $A=8~\text{cm}\cdot 10~\text{cm}=80~\text{cm}^2$
• The new surface area is $204~\text{cm}^2-80~\text{cm}^2=124~\text{cm}^2$.
3. Add the surface areas together to find the total surface area of the 3D object.
• $SA= 224~\text{cm}^2+124~\text{cm}^2$
• $SA=348~\text{cm}^2$