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

7 0

3 years ago

Given 3 sides of a triangle - how to tell it right, acute, or obtuse 8,11, 16

user100 [1]

Using Pythagorean theorem, we can justify our answer.
if c^2 = a^2 + b^2 then triangle is right one,
if c^2 > a^2 + b^2 then triangle is obtuse and
if c^2 < a^2 + b^2 then triangle is acute triangle.

Here a=8, b=11 and c=16
Put these values in the equation
16^2 = 8 ^2 + 11^2
256 = 64 +121
256> 185
So here c^2 > a^2 + b^2 Which means triangle is obtuse triangle.

Answer: Obtuse Triangle

6 0

3 years ago

What is (x+1)(2x-4)(1/x+1)

Taya2010 [7]

(2x^3-6x-4)/x if your simplifying that's the answer

5 0

3 years ago