3 years ago

Which three lengths could be the lengths of the sides of a triangle?

Mathematics

1 answer:

faust18 [17]3 years ago

8 0

B. 10 cm, 15 cm, 24 cm

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Answer:

$Standard\ Form = {3n^2} m^{-2}$

Step-by-step explanation:

Given

$\frac{2}{3m^2n} * 4.5n^3$

Required

Write in Standard Form

To start with; the two monomials have to be multiplied together;

$\frac{2}{3m^2n} * 4.5n^3$

$Standard\ Form = \frac{2 * 4.5n^3}{3m^2n}$

Split the numerator and the denominator

$Standard\ Form = \frac{2 * 4.5 * n^3}{3 * m^2 * n}$

Multiply Like terms

$Standard\ Form = \frac{9 * n^3}{3 * m^2 * n}$

Divide 9 by 3 to give 3

$Standard\ Form = \frac{3 * n^3}{m^2 * n}$

Divide n³ by n to n²

$Standard\ Form = \frac{3 * n^2}{m^2 }$

Split fraction

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$

From laws of indices;

$\frac{1}{a^n} = a^{-n}$

$Standard\ Form = {3 * n^2} * \frac{1}{m^2 }$ becomes

$Standard\ Form = {3 * n^2} * m^{-2}$

Multiply all together

$Standard\ Form = {3n^2} m^{-2}$

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Pls help with this thx

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