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poizon [28]
3 years ago
11

Seven increased by the product of two numbers

Mathematics
1 answer:
Kobotan [32]3 years ago
8 0

Answer: x ≥ 5

Step-by-step explanation:

7 is increased by the product of a number and 3:  7 + 3x

Is at least 22:  ≥ 22

 

Your equation is: 7 + 3x ≥ 22

subtract 7 from both sides

3x ≥ 15

divide both sides by 3

x ≥ 5

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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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