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madreJ [45]
2 years ago
12

4 corners again -9a + 3b - a + 7b - 13 + a - 10 Sorry for the questions pls help

Mathematics
2 answers:
eimsori [14]2 years ago
8 0

Answer:−9a+10b−23

Step-by-step explanation:−9a+3b−a+7b−13+a−10

Combine −9a and −a to get −10a.

−10a+3b+7b−13+a−10

Combine 3b and 7b to get 10b.

−10a+10b−13+a−10

Combine −10a and a to get −9a.

−9a+10b−13−10

Subtract 10 from −13 to get −23.

−9a+10b−23

Tems11 [23]2 years ago
7 0

Answer:

-9a+10b-23

Step-by-step explanation:

-9a+3b-a+7b-13+a-10

-9a-a+a+3b+7b-13-10

(-9a-a+a)+3b+7b-13-10

-9a+(3b+7b)-13-10

-9a+10b-(13+10)

-9z+10b-23

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2. yes; the equation can be written as y = kx; k = 1

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x-y = 0

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If you multiply a number by 1, you get the same thing.

If you multiply x by one, you get y (the same value).

Hope this helps!

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Write a linear factorization of the function. f(x) = x4 + 64x2
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F(X) = x^4 +64x^2 

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3 years ago
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7 0
3 years ago
What is the partial product for 652 x 4= 2,608
Law Incorporation [45]
To find out the answer of the two multiplied products with the help of partial products we separate into unit based numbers so as to, for easier calculation and simplification with units of zeroes. So, in this case we are given the product of 4 times of 652.

Here we need to expand this product of higher number to zeroes and take aside the added numbers to come back to the original multiple of a product. That is:

\huge{652}

\begin{bmatrix}652 \\ + \\ 0 \end{bmatrix}

We just separated and expanded the product to be multiplied by removing other units following it. Same goes for other units at Hundredth, tenth and unitary position.

Therefore,

\begin{bmatrix}600 \\ + \\ 52 \end{bmatrix}

For tenth term.

\begin{bmatrix}50 \\ + \\ 2 \end{bmatrix}

The terms after splitting and expanding the product before multiplication is:

\begin{bmatrix}600 \\ 50 \\ 2 \end{bmatrix}

Multiply the product elements in individual manner and add the elements forged by individual multiplication to get the required solution.

\therefore \quad \begin{bmatrix}600 \times 4 \\ 50 \times 4 \\ 2 \times 4 \end{bmatrix}

\begin{bmatrix}600 \times 4 \\ + \\ 200 \\ + \\ 2 \times 8 \end{bmatrix}

\begin{bmatrix}2400 \\ + \\ 200 \\ + \\ 2 \times 8 \end{bmatrix}

\begin{bmatrix}2600 \\ + \\ 8 \end{bmatrix}

\boxed{\mathbf{Final \: \: Answer \: = \: 2,608}}

Hope it helps.
8 0
3 years ago
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