[Pic of figure] Find the surface area of the prisms

Suman Cole
2021-03-07
Answered

[Pic of figure] Find the surface area of the prisms

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i1ziZ

Answered 2021-03-08
Author has **92** answers

find the area of all sides of rectangle. Then multiply by the number of similar sides by the area. last add all of your answers together.

$6\times 3=18$

$18\times 2=36$

$16\times 3=48$

$48\times 2=96$

$16\times 6=96$

$96\times 2=192$

$36+96+192=324m$

asked 2022-04-30

Simplify the radical $$\sqrt{45}$$ to the simplest form.

asked 2022-04-12

Background

It is relatively easier to find a solution to a system of linear equations in the form of $A\mathbf{\text{v}}=\mathbf{\text{b}}$ given the matrix $A$. But what systematic ways are there that allows us to obtain a matrix given a equation?

For example, consider the following equations with all terms existing in $\mathbb{R}$

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Although it is easy to see that $a=\frac{1}{2},e=\frac{1}{3},i=\frac{1}{4}$ with all other terms being 0 is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = k?

Why I am interested in such question

Consider the vector space ${P}_{2}(\mathbb{R})$, the problem of finding a basis $\beta $ such that $[{x}^{2}+x+1{]}_{\beta}=(2,3,4{)}^{T}$ can be reduced to a problem that has been stated above.

It is relatively easier to find a solution to a system of linear equations in the form of $A\mathbf{\text{v}}=\mathbf{\text{b}}$ given the matrix $A$. But what systematic ways are there that allows us to obtain a matrix given a equation?

For example, consider the following equations with all terms existing in $\mathbb{R}$

$\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]\left[\begin{array}{c}2\\ 3\\ 4\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$

Although it is easy to see that $a=\frac{1}{2},e=\frac{1}{3},i=\frac{1}{4}$ with all other terms being 0 is a viable solution, I am curious if there is a more systematic way of finding a matrix that satisfies a equation. Even more importantly, how should these methods be adapted when there are added constraints on the properties of the matrix? For example, if we require that the matrix of interest should be invertible, or of rank = k?

Why I am interested in such question

Consider the vector space ${P}_{2}(\mathbb{R})$, the problem of finding a basis $\beta $ such that $[{x}^{2}+x+1{]}_{\beta}=(2,3,4{)}^{T}$ can be reduced to a problem that has been stated above.

asked 2021-11-15

To perform: the given operation $0.946\text{}L-210\text{}ml$

asked 2022-04-24

Help with this, please

${x}^{\frac{2}{3}}-3{x}^{\frac{1}{3}}-10=0$

asked 2021-12-10

To evaluate: The simplified form of the expression 2-8.

asked 2022-08-05

$(6{y}^{2}-y-2)\xf7\frac{2y+1}{3y-2}\ufeff=$

asked 2022-05-20

Given a set of n inequalities each of the form ax+by+cz≤d for some a,b,c,d in Q, determine if there exists x, y and z in Q that satisfy all the inequalities.

Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).

I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks

Here is an O(n4) algorithm for solving this: for each triple of inequalities, intersect their corresponding planes to get a point (x,y,z) iff possible. If no such intersecting point exists continue on to the next triple of inequalities. Test each of these intersection points against all the inequalities. If a particular point satisfies all the inequalities the solution has been found. If none of these points satisfy all the inequalities then there is no point satisfying the system of inequalities. There are O(n3) such intersection points and there are n inequalities thus the algorithm is O(n4).

I would like a faster algorithm for solving this (eg: O(n3), O(n2), O(n*logn), O(n)). If you can provide such an algorithm in an answer that would be great. You may notice this problem is a subset of the more general k-dimensional problem where there are points in k-dimensions instead of 3 dimensions as in this problem or 2 dimensions as in my previous problem mentioned above. The time complexity of my algorithm generalized to k dimensions is O(nk+1). Ideally I would like something that is a polynomial time algorithm however any improvements over my naive algorithm would be great. Thanks