Rewrite sin^4xtan^2x in terms of the first power of the cosine

1 answer:

7 0

$\sin ^4 x \tan ^2 x = \dfrac{\sin ^4 x\ \sin ^2 x }{\cos ^2 x} = \dfrac{\sin^6 x}{\cos ^2 x} = \dfrac{(\sin ^2 x)^3}{\cos^2 x}=\dfrac{(1-\cos^2 x)^3}{\cos^2 x}$

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Given: EL- tangent, EK- secant <br> Prove: EJ·LK = EL·LJ

drek231 [11]

Answer:

This is possible.

Step-by-step explanation:

We can say that m<E=m<E, because of the Reflexive Property

Then, we have angles JKL and ELJ, which are equal through the peripheral angle theorem.

With these two angles, we can say that triangles ELK and EJL are similar, by the Angle-Angle Postulate (AA).

Then we can create this ratio through the Corresponding Parts of Similar Triangles Theorem, (CPST), $\frac{LK}{LJ} =\frac{EL}{EJ}$ .

With this ratio, we can cross multiply to get the desired result

$EJ$ · $LK=EL$ · $LJ$

Hope this helps with your RSM problem

Yup, i caught ya.

3 0

3 years ago

Please help thanks so much

Umnica [9.8K]

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$V= \dfrac{4}{3} \pi r^3$

Plug in V=5000π and solve for the radius.

$5000 \pi = \dfrac{4}{3} \pi r^3$

Divide both sides by $\pi$

$5000 = \dfrac{4}{3} r^3$

Multiply both sides by 3/4 (to eliminate the 4/3 on the right side)
Also, flip the equation so we have the unknown variable on the left side (math etiquette).

$r^3=3750$

Take the cube root of both sides

$r \approx 15.5$

Now, calculate the surface area since now we know the radius.
The formula for the surface area is the following:

$S=4 \pi r^2$

Plug in r=15.5

$S \approx 3033$

The second choice is your answer.

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Draw a triangle and you can see that you can use the Pythagorean Theorem.
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