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Rasek [7]
3 years ago
5

Write the slope intercept form of 11x-8y=-48

Mathematics
1 answer:
bija089 [108]3 years ago
8 0
Y=(11/8)x+6
subtract the 8y to the right then add the 48 to the left then divide by 8

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Reduce sixteen over twenty four
yawa3891 [41]
16/24 = 4/6 which = 2/3
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What is the following product? <br>(4x square root 5x^2 + 2^2 square root 6)^2​
tangare [24]

The product is 104 x^{4}+16 \sqrt{30} x^{4}

Explanation:

The given expression is \left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2

Expanding the expression, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)

Now, we shall apply FOIL, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}

Simplifying the terms, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=16 \cdot 5 x^{4}+16 \sqrt{30} x^{4}+4 \cdot 6 x^{4}

Multiplying, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=80 x^{4}+16 \sqrt{30} x^{4}+24 x^{4}

Adding the like terms, we get,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=104 x^{4}+16 \sqrt{30} x^{4}

Thus, the product of the given expression is 104 x^{4}+16 \sqrt{30} x^{4}

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Step-by-step explanation:

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Below.

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Answer:

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4 | | 2 | 5

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- | | 8 |  

| | 2 | 0

| - | 2 | 0

| | | 0:

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