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kiruha [24]
2 years ago
15

Find the value of x for which PQ II RS

Mathematics
1 answer:
Alenkasestr [34]2 years ago
7 0

Answer:

x=3/2

Step-by-step explanation:

\cfrac{3x-1}{21} =\cfrac{2x-2}{12}

36x-12=42x-21

6x=9

x=\cfrac{3}{2}

Hope it helps! :)

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What is the solution set of the equation x2 +3x-4=6
stealth61 [152]
X^2 + 3x - 4 = 6

x^2 + 3x -10 = 0

(x + 5)(x - 2) =0

x = -5 or x = 2

solution set is {-5,2}
6 0
3 years ago
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Answer the question. Show work
RideAnS [48]
We just need to get the perimeter of the backyard.

Whole rectangle: length = 7ft ; width = 5 ft

Perimeter = 2(7+5) = 2(12) = 24 ft


7 0
3 years ago
I NEED HELP ASAP!!!!
dedylja [7]

Answer:

B) 8

Step-by-step explanation:

Triangles MNC and RSC are similar because all 3 angles are the same, so the the ratio between each side should be the same.

If NC=12 and SC=6 then triangle MNC is scaled up by 2 from triangle RSC. This means each side of MNC is 2 times bigger than on RSC. So, sense RS=4, we can multiply that by 2 to get MN=8.

8 0
2 years ago
A rectangular flower garden in Samantha's backyard has 260 feet around its edge. The width of the garden is 60 feet.
OlgaM077 [116]

Answer:

100 - 20 = 80 Length Of the Garden

Step-by-step explanation:

So You Can do 20 Ft Minus 100 Because 100 is the Total so Heres a Graph To Show You

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4 0
2 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}

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7 0
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