What is the answer to this ?

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

3 0

Answer: 13

Step-by-step explanation:

Pythagoras's theorem:

A^2 + B^2 = C^2

12^2 + 5^2 = x^2

144 + 25 = x^2

169 = x^2

13 = x

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2,99,790,000 in scientific notation

ludmilkaskok [199]

So we have 299790000, we want to have only one digit in front of the decimal, 2.9979x10^9

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3 0

3 years ago

Perform the indicated operation:

yaroslaw [1]

The expression $\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }$ is equal to $\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}$ .

<h3>How to simplify a composite algebraic expression</h3>

In this question we must use properties of <em>real</em> algebra to simplify a given expression as most as possible. We proceed to show each step with respective reason.

$\frac{m^{2}-16\cdot n^{2}}{\frac{3\cdot m + 12\cdot n}{m\cdot n} }$ Given
$\frac{m\cdot n \cdot (m^{2}-16\cdot n^{2})}{3\cdot m + 12\cdot n}$ $\frac{\frac{a}{b} }{\frac{c}{d} } = \frac{a\cdot d}{c\cdot b}$
$\frac{m\cdot n \cdot (m-4\cdot n)\cdot (m+4\cdot n)}{3\cdot (m+4\cdot n)}$ Distributive property/ $a^{2}-b^{2} = (a +b)\cdot (a-b)$
$\frac{m\cdot n \cdot (m-4\cdot n)}{3}$ Existence of the multiplicative inverse/Modulative property
$\frac{m^{2}\cdot n}{3} -\frac{4\cdot m\cdot n^{2}}{3}$ Distributive property/ $x^a\cdot x^{b} = x^{a+b}$ /Result