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Jlenok [28]
3 years ago
5

Find the exact value of tan (arcsin (two fifths))

Mathematics
1 answer:
aev [14]3 years ago
5 0
There are 2 ways to do this:

1) Transform tan into terms of sin.
tan x = \frac{sin x}{cos x} = \frac{sin x}{\sqrt{1-sin^2 x}}
where sin x = sin(sin^{-1} (\frac{2}{5})) = \frac{2}{5}
Substituting back in gives:
tan x = \frac{\frac{2}{5}}{\sqrt{1-(\frac{2}{5})^2}} = \frac{\frac{2}{5}}{\sqrt{\frac{21}{25}}} = \frac{2}{5}*\frac{\sqrt{25}}{\sqrt{21}} = \frac{2}{\sqrt{21}}


2) Use a right triangle.
\theta = sin^{-1} (\frac{2}{5}) \\  \\ sin \theta = \frac{2}{5}
sin = opp/hyp --> opp = 2, hyp = 5
Use Pythagorean theorem to solve for adjacent side.
adj = \sqrt{5^2 - 2^2} = \sqrt{21}
tan = opp/adj
tan \theta = \frac{2}{\sqrt{21}}
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Ray Of Light [21]
Okay this is very simple
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7 0
3 years ago
Paula bought a ski jacket on sale for $6 less than half it’s original price.She paid $86 for the jacket.What was the original pr
labwork [276]

Answer: Original price $184

Step-by-step explanation:

(X/2) - 6 = 86

X/2 = 86+6

X/2 = 92

X = 92*2

X = 184

5 0
3 years ago
What is the area of the composite shape
svlad2 [7]
First, let’s find the area of the larger square. We know that the smaller square extend out 2ft and is2ft up. Well take these measures, and subtract the 2ft from the 12 ft at the top, and get 10ft, then add the 2ft to the left side, and find that it’s 6ft. We have these measures: 6ft 6ft 10ft 10ft for the larger rectangle. Let’s find the area. The are formula for a rectangle is Base(10ft) times height(6ft) . Thus 60ftsqrd. Next, let’s do the smaller rectangle. We do the same formula, 4 times 2, and get 8ftsqrd, add them together: 60ft + 8ft = 68ft!
4 0
3 years ago
Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a)
Anna35 [415]

Rolle's Theorem does not apply to the function because there are points on the interval (a,b) where f is not differentiable.

Given the function is f(x)=\sqrt{(2-x^{\frac{2}{3}})^{3}}  and the Rolle's Theorem does not apply to the function.

Rolle's theorem is used to determine if a function is continuous and also differentiable.

The condition for Rolle's theorem to be true as:

  • f(a)=f(b)
  • f(x) must be continuous in [a,b].
  • f(x) must be differentiable in (a,b).

To apply the Rolle’s Theorem we need to have function that is differentiable on the given open interval.

If we look closely at the given function we can see that the first derivative of the given function is:

\begin{aligned}f(x)&=\sqrt{(2-x^{\frac{2}{3}})^3}\\ f(x)&=(2-x^{\frac{2}{3}})^{\frac{3}{2}}\\ f'(x)&=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}\cdot \frac{2}{3}\cdot (-x)^{\frac{1}{3}}\\ f'(x)&=\frac{-\sqrt{2-x^{\frac{2}{3}}}}{\sqrt[3]{x}}\end

From this point of view we can see that the given function is not defined for x=0.

Hence, all the assumptions are not satisfied we can reach a conclusion that we cannot apply the Rolle's Theorem.

Learn more about Rolle's Theorem from here brainly.com/question/12279222

#SPJ4

8 0
2 years ago
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irga5000 [103]
For the first one is has greater then 3 terms
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I believe I hope this helps
4 0
2 years ago
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