Question

I am having a few problems with my homework tonight. 1. f(t)= √ 25-t^2, f(x+5)=√25-x^2+10x+25 is has far as i got!
2. same equation but using f(2x).
3.a(s)= √3 s^2/4 i need to do this with a(2x) and a(3).
4. f(x)=|x|+4 a. f(2),f(-2),f(x^2) and f(x+2) Can anyone help me?

Answer 1

f(x+5)=sqrt25-x^2-10x-25 which if 25 is the only thing under the sqrt then it =5. if the whole equation is under the sqrt then it will be sqrt(-x^2-10x)
clarify before i go onto 2-4

Answer 2

Thw whole thing is under the square root!!

Answer 3

*the

Answer 4

And i dont understand how you got that answer if you dont mind

Answer 5

you had 25-t^2= 25+(-t^2) that should clarify the first part and you replace t with what was in parenthesis.
so
1.sqrt(-x^2-10x)
2.sqrt(25-4x^2)
3.sqrt(3)s/4 sqrt(3)2x/4 sqrt(3)3/4
4.abs(x)+4 (2)and (-2) will be =6 abs(x^2)+4 is x^2+4 abs(x+2)+4= abs(x)+6

Answer 6

so on the first one its not sayin (x+5)(x+5)

Answer 7

it is saying (x+5)(x+5)*-1

Answer 8

okay thanks i got it cause you distributed the negative in the first one and then you cancled the 25 and -25 and is that the final answer

Answer 9

yup

Answer 10

One more question I am not sure how you got the answers for 3.

Answer 11

Wehn i did a(2x)=√3 (2x)^2/4

Answer 12

if it is
then s^2 can be factored out so it is s*sqrt(3)/4