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slega [8]
3 years ago
5

What is the sum of the matrices shown below?

Mathematics
1 answer:
Georgia [21]3 years ago
7 0

Answer:

\red{ \bold{\begin{bmatrix}  - 4 & 26 & - 5  \\ \\6 &  17& - 8\end{bmatrix} }}

Step-by-step explanation:

\begin{bmatrix} 4 & 19 & - 5\\ \\ 7  & 0 & - 14\end{bmatrix} +  \begin{bmatrix}  -8  & 7 & 0\\ \\  - 1  & 17 & 6\end{bmatrix} \\  \\  = \begin{bmatrix} 4 + ( - 8) & 19 + 7 & - 5 + 0\\ \\ 7 + ( - 1)  & 0  + 17& - 14 + 6\end{bmatrix} \\  \\ = \begin{bmatrix} 4  - 8 & 19 + 7 & - 5 + 0\\ \\ 7  - 1  & 0  + 17& - 14 + 6\end{bmatrix} \\  \\ \purple{ \bold{ = \begin{bmatrix}  - 4 & 26 & - 5  \\ \\6 &  17& - 8\end{bmatrix} }}

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In the following figure, m∠A = 55° and the m∠ACE = 86° . What is the m∠E = ? Show your work.
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A triangle has side lengths of 33,56, and 65. Is it a right triangle? Explain.
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How do you do number 1?<br>Whenever I tried to answer it, I always get fraction. help me.​
Flura [38]

Answer:

The pairs are (13,15) and (-15,-13).

Step-by-step explanation:

If n is an odd integer, the very next odd integer will be n+2.

n+1 is even (so we aren't using this number)

The sum of the squares of (n) and (n+2) is 394.

This means

(n)^2+(n+2)^2=394

n^2+(n+2)(n+2)=394

n^2+n^2+4n+4=394               since (a+b)(a+b)=a^2+2ab+b^2

Combine like terms:

2n^2+4n+4=394

Subtract 394 on both sides:

2n^2+4n-390=0

Divide both sides by 2:

n^2+2n-195=0

Now we need to find two numbers that multiply to be -195 and add up to be 2.

15 and -13 since 15(-13)=-195 and 15+(-13)=2

So the factored form is

(n+15)(n-13)=0

This means we have n+15=0 and n-13=0 to solve.

n+15=0

Subtract 15 on both sides:

n=-15

n-13=0

Add 13 on both sides:

n=13

So if n=13 , then n+2=15.

If n=-15, then n+2=-13.

Let's check both results

(n,n+2)=(13,15)

13^2+15^2=169+225=394.  So (13,15) looks good!

(n,n+2)=(-15,-13)

(-15)^2+(-13)^2=225+169=394.  So (-15,-13) looks good!

6 0
3 years ago
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