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3 years ago

What is the 9th multiple of 15

Mathematics

1 answer:

MArishka [77]3 years ago

5 0

135 because
15x1=15 15x7=105
15x2=30 15x8=120
15x3=45 15x9=135
15x4=60 15x10=150
15x5=75 15x11=165
15x6=90 15x12=190

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ale4655 [162]

N/6=30
Multiply both sides by 6
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2 years ago

Difference between standard deviation and mean deviation

Yakvenalex [24]

$Question$

difference between standard deviation and mean deviation

Answer:

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3 years ago

Complete the factorization of 4x2 – 9.

Nastasia [14]

Your Answer would be:

4x2<span> – 9 = (2x - 3)(2x + 3)
</span>
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4 0

3 years ago

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What is a factor of 17

eduard

A factor of 17 is 1. Another is 17.

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3 years ago

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Which expression is equivalent to sqrt8x^7y^8? assume x>0

Ilia_Sergeevich [38]

Answer:

$\sqrt{8x^7y^8$ $= 2x^3y^4 * \sqrt{2x}$

Step-by-step explanation:

Given

$\sqrt{8x^7y^8$

Required

Evaluate

$\sqrt{8x^7y^8$

Separate each factor with multiplication sign

$\sqrt{8*x^7*y^8$

Express 8 as 4 * 2 and $x^7$ as $x^6 * x$

$\sqrt{4 * 2 *x^6 * x*y^8$

Split each factor

$\sqrt{4} * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}$

Evaluate $\sqrt4$

$2 * \sqrt{2} *\sqrt{x^6} * \sqrt{x}*\sqrt{y^8}$

Apply law of indices:

$2 * \sqrt{2} * x^{6*\frac{1}{2}} * \sqrt{x} * y^{8*\frac{1}{2}}$

$2 * \sqrt{2} * x^3 * \sqrt{x} * y^4$

Reorder

$2* x^3 * y^4 * \sqrt{2} * \sqrt{x}$

$2x^3y^4 * \sqrt{2} * \sqrt{x}$

$2x^3y^4 * \sqrt{2*x}$

$2x^3y^4 * \sqrt{2x}$

$2x^3y^4 \sqrt {2x$

Hence:

$\sqrt{8x^7y^8$ $= 2x^3y^4 * \sqrt{2x}$

8 0

3 years ago

Read 2 more answers