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3 years ago

2.5(10 times+8)+3times

Mathematics

2 answers:

balandron [24]3 years ago

5 0

Answer:

28x+20

Step-by-step explanation:

2.5 times 10x= 25x

2.5times 8 = 20

25x+3=28

28x+20

MakcuM [25]3 years ago

4 0

2.5(10x+8)+3x

Use distributive property first:

25x + 20 + 3x

Combine like terms:

28x + 20

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A cube has a volume of 8 cubic inches. what is the length of each edge of the cube?

snow_tiger [21]

$2 \times 2 \times 2$

5 0

3 years ago

Read 2 more answers

-112=16e Solve for e

snow_tiger [21]

Answer: -7 = e

Step-by-step explanation: To solve this equation for e, we need to get e by itself on the right side of the equation. Since e is multiplied by 16, in order to get e by itself, we need to divide by 16 on the right side of the equation. If we divide by 16 on the right side of the equation, we must also divide by 16 on the left side of the equation.

On the right side of the equation the 16's cancel and we have e. On the right side of the equation -112 divided by 16 is -7. Remember that a negative divided by a positive is a negative.

So we have -7 = e which is the solution to our equation.

To check our solution, we can plug -7 in for e in the original equation. So we have -112 = 16 (-7) or -112 = -112 which is a true statement so our answer checks.

5 0

3 years ago

Plz help urgent !!!! will give the brainliest !!

Mkey [24]

Answer:

$X = \begin{bmatrix}1&3\\ 2&4\end{bmatrix}$

Step-by-step explanation:

The question we have at hand is, in other words,

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}\left(X\right)=\begin{bmatrix}8&20\\ \:3&7\end{bmatrix}$ - where we have to solve for the value of $X$

If we have to isolate $X$ here, then we would have to take the inverse of the following matrix ...

$\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}$ ... so that it should be as follows ... $\begin{bmatrix}4&2\\ \:1&1\end{bmatrix}^{-1}$

Therefore, we can conclude that the equation as to solve for " $X$ " will be the following,

$X=\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ - First find the 2 x 2 matrix inverse of the first portion,

$\begin{bmatrix}4&2\\ 1&1\end{bmatrix}^{-1}$ = $\frac{1}{\det \begin{pmatrix}4&2\\ 1&1\end{pmatrix}}\begin{pmatrix}1&-2\\ -1&4\end{pmatrix}$ = $\frac{1}{2}\begin{bmatrix}1&-2\\ -1&4\end{bmatrix}$ = $\begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}$

At this point we have to multiply the rows of the first matrix by the rows of the second matrix,

$X = \begin{bmatrix}\frac{1}{2}&-1\\ -\frac{1}{2}&2\end{bmatrix}\begin{bmatrix}8&20\\ 3&7\end{bmatrix}$ ,

$X = \begin{pmatrix}\frac{1}{2}\cdot \:8+\left(-1\right)\cdot \:3&\frac{1}{2}\cdot \:20+\left(-1\right)\cdot \:7\\ \left(-\frac{1}{2}\right)\cdot \:8+2\cdot \:3&\left(-\frac{1}{2}\right)\cdot \:20+2\cdot \:7\end{pmatrix}$ - Simplifying this, we should get ...