49q²-70q+25 resolve into fractors​

(7q-5)²

Step-by-step explanation:

= 49q² - 70q + 25

= 49q² -35q -35q + 25

= 7q(7q-5) - 5(7q - 5)

= (7q-5) (7q-5)

= (7q-5)²

CMIIW

Step-by-step explanation:

49q² -70q +25 , rewrite the first two terms as squares

(7q)² -70q + 5² , compare with  a² - 2ab + b² = (a-b)²

    a² - 2ab + b²

(7q)² -2(7q)(5 )+ 5², because 70q = 2*35q = 2*7*5*q

We can see that a = 7q and b=5, therefore

49q² -70q +25 = (7q -5)²

We set the equation equal to 0 to find the roots

7q-5 =0, add 5 to both sides

7q =5 , divide both sides by 7

q = 5/7 ≈ 0.714

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

can you rephrase the question ? it is unclear : -)

step-by-step explanation: