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vichka [17]
3 years ago
15

PLEASE HELP FOR 87 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Anton [14]3 years ago
6 0
1. You can expand the coefficients to yield 6, and add the powers since the bases are the same (ie A)

2. y⁻² can also be rewritten as 1/y², so it becomes 1/y² · 1/y⁵ = 1/y⁷ (ie C)

3. Only the x is affected by the power.

4. y = 3·5ˣ doesn't change in slope, so it essentially is similar to a 5ˣ graph.

5. Should be pretty straight forward.
6. Should be pretty straight forward.
7. Should be pretty straight forward.

8. Since x is the length of the border, calculate the area of the frame and subtract that from the area of the square painting. (Remember that the length of the square painting is x - 8, not x - 4)

9. Should be pretty straight forward.
11 - 25 are pretty much similar to each other.
If you have trouble, let me know and I'll walk you through it.
ivanzaharov [21]3 years ago
3 0

Answer:

x=9

x=2

Step-by-step explanation:

Calculate the area of the frame and subtract that from the area of the square painting. (Remember that the length of the square painting is x - 8, not x - 4)

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For waht values of x do the vectors -1,0,-1), (2,1,2), (1,1, x) form a basis for R3?
DaniilM [7]
<h2>Answer:</h2>

The values of x for which the given vectors are basis for R³ is:

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<h2>Step-by-step explanation:</h2>

We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.

We are given three vectors as:

(-1,0,-1), (2,1,2), (1,1, x)

The matrix formed by these vectors is:

\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]

Now, the determinant of this matrix is:

\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-1(x-2)-2(1)+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+2-2+1\\\\\\\begin{vmatrix}-1 &2 & 1\\ 0& 1 & 1\\ -1 & 2 & x\end{vmatrix}=-x+1

Hence,

-x+1\neq 0\\\\\\i.e.\\\\\\x\neq 1

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