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3 years ago

Use addition to solve the linear system of equations. Include all of your work in your final answer.

1 answer:

4 0

We have that
<span>x - y = 4-------> multiply by -1-----> -x+y=-4-----> equation 1
x+ 2y = 4------> equation 2

add equation 1 and equation 2
-x+y=-4
x+2y=4
-----------------
0+3y=0-----> y=0
x+2y=4------> x+0=4------>x=4

the solution is the point (4,0)</span>

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Which of the following is the solution to the inequality below ? -6/5(4-x)<_-4(X+1/2)

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Answer:

The answer is the option D

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Step-by-step explanation:

we have

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3 years ago

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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3 years ago