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3 years ago

What is the answer to 2[n+6]=80

Mathematics

2 answers:

makvit [3.9K]3 years ago

5 0

N=34 Would be your answer.

Gre4nikov [31]3 years ago

4 0

2[n+6]=80 +3,6-(8)-4,5 lol I'm so smart good thing my mom taught me well

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How many terms are in the expression below?<br> 17 – (5n = 4) + 8%

tankabanditka [31]

Answer: 4

Step-by-step explanation:

(Can someone please answer my math question)

7 0

3 years ago

Use matrix addition to solve this equation: B + 15 −7 4 0 1 2 = 1 2 12 4 0 2 b11 = b12 = b13 = b21 = 4 b22 = −1 b23 = 0

Arada [10]

Answer:

$b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$

Step-by-step explanation:

The given matrix addition is

$B+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

We need to find the elements of matrix B.

Let $B=\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}$

Substitute the value of matrix.

$\begin{bmatrix}b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\end{bmatrix}+\begin{bmatrix}15&-7&4\\ 0&1&2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

After addition of two matrix we get

$\begin{bmatrix}b_{11}+15&b_{12}-7&b_{13}+4\\ b_{21}+0&b_{22}+1&b_{23}+2\end{bmatrix}=\begin{bmatrix}1&2&12\\ 4&0&2\end{bmatrix}$

On equating both sides.

$b_{11}+15=1\Rightarrow b_{11}=-14$

$b_{12}-7=2\Rightarrow b_{12}=9$

$b_{13}+4=12\Rightarrow b_{13}=8$

$b_{21}+0=4\Rightarrow b_{21}=4$

$b_{22}+1=0\Rightarrow b_{22}=-1$

$b_{23}+2=2\Rightarrow b_{23}=0$

Therefore, the elements of matrix B are $b_{11}=-14,b_{12}=9,b_{13}=8,b_{21}=4,b_{22}=-1,b_{23}=0$ .

3 0

3 years ago

A car is traveling at 47 mph. If its tires have a diameter of 27 inches, how fast are the car's tires turning? Express the answe

Amiraneli [1.4K]

Answer:

3676.44 rad/min

Step-by-step explanation:

It is a problem about the angular speed of the car's wheel.

You can calculate the angular speed by using the following formula, which relates the tangential speed of the wheels (the same as the speed of the car) with the angular speed:

$\omega=\frac{v}{r}$ ( 1 )