√4x²-20x+25=5-2x

HELP PLEASE!!! P. S the entire equation 4x²-20x+25 is under the square root

The equal is valid for x≤ 5/2

Step-by-step explanation:

sqrt(4x²-20x+25)=5-2x

Square both sides

(sqrt(4x²-20x+25))^2=(5-2x)^2

Foil the term on the right (5-2x)(5-2x) = 25 -10x-10x+4x^2 = 25 -20x+4x^2

(4x²-20x+25)=25-20x+4x^2

The left side is equal to the right side

We need to check for extraneous solutions

The break point is 5-2x =0

5 =2x

x =5/2

Check x < 5/2

Let x=0

sqrt(4x²-20x+25)=5-2x

sqrt(25)=5

True

Check x=5/2

sqrt(4x²-20x+25)=5-2x

sqrt(4(25/4)-20*5/2+25)=5-2(5/2)

sqrt(25-50+25)=5-5

0=0

True

Check x>5/2

Let x=10

sqrt(4x²-20x+25)=5-2x

sqrt(4*100-20*10+25)=5-2*10

sqrt(400-200+25)=5-20

sqrt(225) = -15

15 does not equal -15

False

The equal is valid for x≤ 5/2

3, 12, 48, 384, 1536, 6144, 24576, 98304

step-by-step explanation:

just keep timing it by 4.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: