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3 years ago

Help me, please ?? :)

Mathematics

1 answer:

oee [108]3 years ago

5 0

Answer:

a) 11

b) 16

c) between 5 and 6

d) 16

Step-by-step explanation:

$\text{a. }\quad\sqrt{121}=\sqrt{11^2}=\boxed{11}\\\\\text{b. }\quad 8\sqrt{4}=8\sqrt{2^2}=8\cdot 2=\boxed{16}\\\\\text{c. }\quad\sqrt{35}\ \dots\ \sqrt{25}$

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3 years ago

PLEASE HELP WILL GIVE BRAINLIEST! What is the following sum? Assume x > or equal to 0 and y is > or equal to 0.

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Answer:

Option (b) is correct.

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Step-by-step explanation:

Given:Expression $\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

We have to find the sum of the given expression and choose the correct from the given options.

Consider the given expression $\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$ .

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Now, the given expression becomes,

$\sqrt{x^2y^3}+2\sqrt{x^3y^4}+xy\sqrt{y}$

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3 years ago

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Step-by-step explanation:

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