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

7 0

3 years ago

How can i learn times faster like 3's til 12's in Times i have a hard time in my times can someone tell me how i can learn them

iogann1982 [59]

To get better at 12's:
Write down on your paper your 1's facts in column skip 5 and 11 going to 14 (a vertical line - line that goes up and down). To the right of that column, write your two's facts 0 to 8 and repeat again. Then you will have your 12's! Should look as follows

12's:

0 0 = 12 x0
1 2 = 12 x1
2 4 = 12 x2
3 6 = 12 x3
4 8 = 12 x4
6 0 = 12 x5
7 2 = 12 x6
8 4 = 12 x7
9 6 = 12 x8
10 8 = 12 x9
12 0 = 12 x10
13 2 = 12 x 11
14 4 = 12 x 12

5 0

3 years ago

-15 Assignment<br> 5x2=<br> 3

vladimir2022 [97]

Not really sure what you’re saying; but 5x2=10

6 0

3 years ago