Find the value of x if PQ = x^2 + 3, QR = 4 + 2x, and PR = 15 if Q is between P and R.

x=2

Step-by-step explanation:

PQ+QR=PR

=x^2+3+4+2x=15

=x^2+2x+7=15

=x^2+2x+7-15=0

=x^2+2x-8=0. (quadratic equation)

=(x+4)(x-2)=0. (factorise)

=x+4=0 or x-2=0

x=-4 or 2

we take x to be the positive value which is

x=2

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.