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3 years ago

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Factor a^2b^2-7ab+10

1 answer:

pentagon [3]3 years ago

5 0

What you should do in this case is to see the terms that are repeated in the expression.
We have then that the terms are a and b.
Making a common factor we have:
a * b (a * b -7) +10
Answer:
The equivalent expression is:
a * b (a * b -7) +10

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2 tan 30°<br>II<br>1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

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MAVERICK [17]

Divide 7 into 5 to find each penny value.
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