All categories

3 years ago

The volume of a storage pod that is shaped like a cube is 1728 cubic feet what is the edge length of the storage pod?

Mathematics

1 answer:

UNO [17]3 years ago

5 0

Answer:

Step-by-step explanation:

a=V^1/3

a=1728^1/3 or a= $\sqrt[3]{1728}$

12*12*12=1728

You might be interested in

Which statement is true? 1) The y-value of Function A when x=-1 is greater than the y-value of Function B when x=-1. 2) The y-va

labwork [276]

The answer is 2 I believe

7 0

3 years ago

Read 2 more answers

mia hired a moving company, it charged 500$ for service mia give them a 16% tip.How much did she tip the movers,

snow_tiger [21]

Answer: 80 dollars

Step-by-step explanation:

You can find a percentage with one of two ways.

For hundreds like this, oftentimes it's quicker and more simple to think of it as just 100. The proportion from 100 to 500 is just 5. 16% of 100 is 16. So, multiply 16 * 5. You would get 80.

The second method is you can use 0.16 * 500. It's the more conventional method, and you get 80, like the last method.

3 0

3 years ago

Write expression using the distributive property to find the product of 9 x82

irga5000 [103]

Answer:

Step-by-step explanation:

9 * 82 = ( 10-1) * 82

= 10*82 - 1 *82

= 820 - 82

= 738

5 0

3 years ago

Ac=22 bc=x+14 ab=x+10 find x

Nesterboy [21]

AB + BC = AC Segment Addition Postulate

(x + 10) + (x + 14) = 22 Substitution

2x + 24 = 22 Simplify (added like terms)

2x = -2 Subtraction Property of Equality

x = -1 Divison Property of Equality

Answer: x = -1

5 0

3 years ago

) a, p and d are n×n matrices. check the true statements below:

BigorU [14]

A. False. Consider the identity matrix, which is diagonalizable (it's already diagonal) but all its eigenvalues are the same (1).

b. True. Suppose $\mathbf P$ is the matrix of the eigenvectors of $\mathbf A$ , and $\mathbf D$ is the diagonal matrix of the eigenvalues of $\mathbf A$ :

$\mathbf P=\begin{bmatrix}\mathbf v_1&\cdots&\mathbf v_n\end{bmatrix}$

$\mathbf D=\begin{bmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{bmatrix}$

Then

$\mathbf{AP}=\begin{bmatrix}\mathbf{Av}_1&\cdots&\mathbf{Av}_n\end{bmatrix}=\begin{bmatrix}\lambda_1\mathbf v_1&\cdots&\lambda_n\mathbf v_n\end{bmatrix}=\mathbf{PD}$

In other words, the columns of $\mathbf{AP}$ are $\mathbf{Av}_i$ , which are identically $\lambda_i\mathbf v_i$ , and these are the columns of $\mathbf{PD}$ .

c. False. A counterexample is the matrix

$\begin{bmatrix}1&1\\0&1\end{bmatrix}$

which is nonsingular, but it has only one eigenvalue.

d. False. Consider the matrix

$\begin{bmatrix}0&1\\0&0\end{bmatrix}$

with eigenvalue $\lambda=0$ and eigenvector $\begin{bmatrix}k&0\end{bmatrix}^\top$ , where $k\in\mathbb R$ . But the matrix can't be diagonalized.

7 0

3 years ago