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aleksley [76]
3 years ago
8

First, rewrite 9/10 and 23/25 so that they have a common denominator.

Mathematics
2 answers:
alekssr [168]3 years ago
3 0
Multiply the numerator and denominator of 23/25 by 10 and do the same on 9/10 but multiply by 25. Your denominator will be 250. You can figure the rest out.
Assoli18 [71]3 years ago
3 0
Well you have to have the common at 50 because that it so multiply23/25 by 2 and 9/10 by 5 so 45/50 and 46/50
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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}

7 0
4 years ago
How to simplify 12 over 16
Sveta_85 [38]
12/16 can be simplified to 3/4, because 12 and 16 can both be divided 4, 12/4=3, 16/4=4.
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