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patriot [66]
3 years ago
14

7(m+5) = 21 What is it?​

Mathematics
2 answers:
pav-90 [236]3 years ago
7 0

Answer:

m = -2

Step-by-step explanation:

Look at the attached image. First, we divide both sides by 7 to get m + 5 = 3. Now, we subtract 5 from both sides to get m = -2. Now, we substitute m for -2. As we can see, it checks out.

son4ous [18]3 years ago
6 0
M= -2 because you divide both side by 7 to get m+5=3 then subtract 5 from both sides to get m=-2
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F(x) = x5 + (x + 3)2 x=-1
sveticcg [70]
Hello!

You put -1 in for x

F(-1) = (-1)^5 + (-1 + 3)^2

F(-1) = -1 + (2)^2

F(-1) = -1 + 4

F(-1) = 3

The answer is 3

Hope this helps!
8 0
3 years ago
What is being constructed based on the markings in the following diagram???
mamaluj [8]
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MariettaO [177]

Answer:

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Step-by-step explanation:

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5 0
3 years ago
I need help fast plzzzzz
PolarNik [594]

Answer:

68

Step-by-step explanation:

Find the vanilla first by dividing 420 by 7 and multiplying the quotient by 2 to get 120.

Then find the banana using the equation 35/100 = x/420, cross multiply to get 147.

Add 120 and 147 to get 267.

Subtract 267 from 420 to get 153.

Divide 153 by 9 to get 17.

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4 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
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