Question

Use basic identities to simplify the expression.
cscx cosx/ secx

Answer 1

\[\csc x \cot x/\sec x\]

Answer 2

\[\Large\frac{\csc x}{\sec x}\cot x = \frac{\frac{1}{\sin x}}{\frac{1}{\cos x}}\cot x = \frac{\cos x}{\sin x}\cot x= \cot x \cot x = \cot^2 x\]

Answer 3

can anyone help me? The centripetal acceleration of an object moving uniformly in a circle varies inversely with the radius of the circle. If the object feels acceleration of 20 m/s2 when the radius is 4 m, find the acceleration when the radius is 5 m.

Answer 4

@allisonxloveyx
\[\Huge \color{red}{a_c = \frac{v^2}{r}}\]

Answer 5

Thanks for invading my question... @allisonxloveyx

Answer 6

D: Sorry I really need hep on this. :/ I've been trying to figure it out for an hour.

Answer 7

Well, do you understand how to solve this problem now?

Answer 8

kinda yes kinda no. The formula I was given by my teacher is I~1/d^2 so then you get I=k/d^2

Answer 9

Never see formula like that. Do you know what each variable represents?

Answer 10

\[\frac{ \csc x . \cot x }{ \sec x } \]
\[\frac{ \frac{ 1 }{ \sin x }.\frac{ \cos x }{ \sin x } }{ \frac{ 1 }{ \cos x } }\]
\[\frac{ 1 }{ \sin x } .\frac{ \cos x }{ \sin x } . \cos x\]
\[\frac{ \cos ^{2}x }{ \sin ^{2}x }\]
\[\cot ^{2}x\]

Answer 11

Well the problem was for at 5m the flashlight illumates at 70 cadnera. So I just need to figure out what variables to switch around to help it fit my problem.