How to solve the following system y=(1/2)x^2+2x-1 and 3x-y=1

(0,-1),(2,5)

Step-by-step explanation:

3x-y=1

0.5+2x-1=y

Substitute the equation of y into the other equation

3x-(0.5+2x-1)=1

3x-0.5-2x+1=1

collect like terms

-0.5+x=0

Solve for x

x(-0.5x+1)=0

0=-0.5x+1

-1=-0.5x

x=2 x=0

Back substitute into the other original equation and solve for y

3(2)-y=1

6-y=1

y=5

(2,5)

3(0)-y=1

-1(-y)=-1(1)

y=-1

(0,-1)

what are the options? ?

step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length