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Shtirlitz [24]
3 years ago
14

Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Mathematics
1 answer:
Hitman42 [59]3 years ago
7 0
\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\

\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}

\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}

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HACTEHA [7]
Distance formula
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3 years ago
Find the area of the region which is inside the polar curve r=5sin(θ) but outside r=4. Round your answer to four decimal places
natka813 [3]

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be 3.75 square units.

<h3>What is an area bounded by the curve?</h3>

When the two curves intersect then they bound the region is known as the area bounded by the curve.

The area of the region which is inside the polar curve r = 5 sinθ but outside r = 4 will be

Then the intersection point will be given as

\rm 5 \sin \theta  = 4\\\\\theta = 0.927 , 2.214

Then by the integration, we have

\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214}[ (5 \sin \theta)^2 - 4^2] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [25\sin ^2 \theta - 16] d\theta \\\\\\\rightarrow \dfrac{1}{2} \times \int _{0.927}^{2.214} [ \dfrac{25}{2}(1 - \cos 2\theta ) - 16] d\theta \\

\rightarrow \dfrac{1}{2} [\dfrac{25 \theta }{5} - \dfrac{25 \cos 2\theta }{2} - 16\theta]_{0.927}^{2.214} \\\\\\\rightarrow \dfrac{1}{2} [\dfrac{25(2.214 - 0.927) }{5} - \dfrac{25 (\cos 2\times 2.214 - \cos 2\times 0.927) }{2} - 16(2.214 - 0.927]\\

On solving, we have

\rightarrow \dfrac{1}{2} \times 7.499\\\\\rightarrow 3.75

Thus, the area of the region is 3.75 square units.

More about the area bounded by the curve link is given below.

brainly.com/question/24563834

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5 0
2 years ago
Which set of shapes could be used to form a net for a square pyramid?
bulgar [2K]

Answer:

A.1 square and 4 triangles

Step-by-step explanation:

A pyramid has sides that are triangular faces and a base. In a square pyramid, the base is a square.

The net therefore has 1 Square and 4 Triangles.

A net is given in the attached diagram:

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To eliminate the x terms and solve for y in the fewest steps, by which constants should the equations be multiplied by before ad
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Step-by-step explanation:

4 0
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Read 2 more answers
25 POINTS PLEASE HELP ME!! Sam conjectures that for x ≤ - 2, it is true that x^5 + 7 &gt; x^3. Is Sam’s conjecture correct? Why
CaHeK987 [17]

The true statement about Sam’s conjecture is that the conjecture is not correct

<h3>How to determine if Sam’s conjecture is correct or not?</h3>

Sam’s conjecture is given as:

For x ≤ - 2

It is true that x^5 + 7 > x^3.

The inequality x ≤ - 2 means that the highest value of x is -2

Assume the value of x is -2, then we have:

(-2)^5 + 7 > (-2)^3

Evaluate the exponents

-32 + 7 > -8

Evaluate the sum

-25 > -8

The above inequality is false because -8 is greater than -25 i.e. -8 > -25 or -25 < -8

Hence, the true statement about Sam’s conjecture is that the conjecture is not correct

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5 0
1 year ago
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