Match the equations with their solutions over the interval [0, 2π].

Step-by-step explanation:

A). cos(x)tan(x) =

cos(x) =

sin(x) =

Since sine is positive in 1st and 2nd quadrant.

Therefore, x =

B). sec(x)cot(x) + 2 = 0

Since sine is negative in 3rd and 4th quadrant.

Therefore, x = and

C). sin(x)cot(x) + = 0

cosx = -

Since cosine is negative in 2nd and 3rd quadrant.

Therefore, x = and

D). csc(x)tan(x) - 2 = 0

cos(x) =

Since x is positive in 1st and 4th quadrant.

x = and

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.