Lisa bucket weighs 21kg after pouring out half it weigh 12kg what’s the wight of the bucket?

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

2 years ago

Hey everyone hru and i hope u all are doing good

Elan Coil [88]

Answer:

Pacific question or no because if not we can't answer it but I'm doing good

6 0

2 years ago

Read 2 more answers

Help me please please

Vera_Pavlovna [14]

Answer:

3.88 nm

Step-by-step explanation:

The angle at Port Latta between the actual direction of travel and the destination at Lookout Point is 90° -75° = 15°. The sine function relates the angle, the side opposite, and the hypotenuse of a right triangle, so you have ...

sin(15°) = (distance to Lookout Point)/(15 nm)

Solving for the desired distance, we have ...

distance to Lookout Point = (15 nm)sin(15°) = 3.88 nm

The yacht is 3.88 nautical miles from Lookout Point.

3 0

2 years ago

On a snow day, Owen created two snowmen in his backyard. Snowman A was built to

faust18 [17]

Answer:

The equations for each function:

A(t) = 60 - 6t
B(t) = 42 - 3t

The lines plotted, see attached

The point at which both the snowmen have same height:

60 - 6t = 42 - 3t
6t - 3t = 60 - 42
3t = 18
t = 6

The required interval is

t = [0, 6)

7 0

2 years ago

R/3+2=21 solve for r

olchik [2.2K]

Answer:

r=63

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

r/3-(21)=0 r

Simplify —

3

r

— - 21 = 0

3 2.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 3 as the denominator :

21 21 • 3

21 = —— = ——————

1 3

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

7 0

3 years ago