if triangle NPQ is congruent to triangle RSQ the perimeter of triangle NPQ= 7x+2, and the perimeter of triangle RSQ=10x-4, find the perimeter of triangle NPQ

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

happy to

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if y=4x then you will substitute 4x in for y in the second equation.

8x + 4x = 18 combine like terms

12x = 18 divide by 12 on both sides

x = 18/12 or 3/2

now replace x in the first equation with 3/2 to get

y = 4 3/2

y = 6

the answer is a

(3/2,6)

hope that .

brainliest is always apprectiated

f(x) is   are being squared

g(x) is   increasing by 2

h(x) is adding .25 each time

k(x) is taking the square root

f(x) and k(x) are exponential functions