Find the exact value of csc theta for an angle theta with cosine theta with equals -2/5 and with its terminal side in quadrant 2

A

Step-by-step explanation:

Cos(theta) = -2/5

From sin^2(theta) + cos^2(theta) = 1 we get

sin^2(theta) = 1 - cos^2(theta)

sin(theta) = sqrt(1 - cos^2(theta)). The root is the positive one in quad 2

sin(theta) = sqrt(1 - cos^2(theta))

sin(theta) = sqrt(1 - (-2/5)^2)

sin(theta) = sqrt(1 - 4/25)

sin(theta) = sqrt(21/25)

sin(theta) = sqrt(21)/5

csc(theta) = 1/sin(theta)

csc(theta) = 1/sqrt(21)/5

csc(theta) = 5/sqrt(21)

The answer should be the denominator rationalized as in

csc(theta) = 5 * sqrt(21) / 21.

However that is not an option, so the answer is A

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )