Question

Can anyone help me out? got some algebra 2..

Answer 1

What is the solution set to the following set?

x+y=4
x^2 + y^2=16

Answer 2

see its x + y = 4
on squaring both sides
\[\(x \+\ y)\^\2\ \=\ \1\6\\]
so
x^2 + y^2 + 2xy = 16

Answer 3

and u r given tht x^2 + y^2 = 16 i.e the second equation
so u get 2xy = 0

Answer 4

@Cman456

Answer 5

(4,0) (0,-4)?

Answer 6

and wht about (4,0)
hey it can't be negative,......its given x + y =4

Answer 7

sorry i meant (0,4)

Answer 8

so( 4,0) (0,4)

Answer 9

yeah.....

Answer 10

Another one?

Answer 11

Y=(x-3)(x+2)(x-2) what are the zeroes of the function?

Answer 12

-3,2.-2?

Answer 13

its supposed to be 3, -2, 2

Answer 14

um why

Answer 15

nope nvm got it

Answer 16

another one please haha

Answer 17

classify 4x^5 + 2x^4 -5x^3+12 by number of terms

Answer 18

polynomial of 4 terms?

Answer 19

or trinomial?

Answer 20

4 terms.