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Marina86 [1]
3 years ago
14

le="3 \sqrt{2} - 4 \div \sqrt{3} - 2" alt="3 \sqrt{2} - 4 \div \sqrt{3} - 2" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
ivann1987 [24]3 years ago
4 0
Simplify the following:
(3 sqrt(2) - 4)/(sqrt(3) - 2)

Multiply numerator and denominator of (3 sqrt(2) - 4)/(sqrt(3) - 2) by -1:
-(3 sqrt(2) - 4)/(2 - sqrt(3))

-(3 sqrt(2) - 4) = 4 - 3 sqrt(2):
(4 - 3 sqrt(2))/(2 - sqrt(3))

Multiply numerator and denominator of (4 - 3 sqrt(2))/(2 - sqrt(3)) by sqrt(3) + 2:
((4 - 3 sqrt(2)) (sqrt(3) + 2))/((2 - sqrt(3)) (sqrt(3) + 2))

(2 - sqrt(3)) (sqrt(3) + 2) = 2×2 + 2 sqrt(3) - sqrt(3)×2 - sqrt(3) sqrt(3) = 4 + 2 sqrt(3) - 2 sqrt(3) - 3 = 1:
((4 - 3 sqrt(2)) (sqrt(3) + 2))/1

((4 - 3 sqrt(2)) (sqrt(3) + 2))/1 = (4 - 3 sqrt(2)) (sqrt(3) + 2):
Answer: (4 - 3 sqrt(2)) (sqrt(3) + 2)
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Which is an equation of a direct proportion?
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2 years ago
Suppose that the terminal side of angle alphaα lies in Quadrant I and the terminal side of angle betaβ lies in Quadrant IV. If s
melamori03 [73]

Solution :

It is given that :

$\alpha$ lies in the first quadrant.

And $\beta$ lies in the fourth quadrant.

Since, $\sin \alpha = \frac{5}{13}$     and $\cos \beta = \frac{6}{\sqrt{85}}$    (given)

$\sin \alpha = \frac{5}{13}$  

$\cos \alpha = \sqrt{1-\sin^2 \alpha}$

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Similarly  $\cos \beta = \frac{6}{\sqrt{85}}$

$\sin \beta = \sqrt{1-\cos^2 \beta}$

$\sin \beta = \sqrt{1-\frac{36}{85}}$

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Therefore,

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3 years ago
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