Find all the points if any where y-4x=12 intersects 2-y=2(x+2)^2 a) there are no points of intersection b) (-3,0) c) (3,0) d) (-3,0) and (3,0)

B

Step-by-step explanation:

given the 2 equations

y - 4x = 12 → (1)

2 - y = 2(x + 2)² → (2)

rearrange (1) expressing y in terms of x

y = 12 + 4x

Simplify (2) by expanding factor and substituting y = 12 + 4x

2 - (12 + 4x) = 2(x² + 4x + 4)

2 - 12 - 4x = 2x² + 8x + 8

- 10 - 4x = 2x² + 8x + 8 ← rearrange into standard form

add 10  + 4x to both sides

2x² + 12x + 18 = 0 ← in standard form

divide through by 2

x² + 6x + 9 = 0

(x + 3)(x + 3) = 0 ⇒ (x + 3)² = 0 ⇒ x = - 3

Point of intersection = (- 3, 0 ) → B

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )

(b) 240

step-by-step explanation:

first, we use the pythagorean formula using the given side lengths to find x. then we use x to find the side lengths. then we add the side lengths to find the perimeter.

pythagorean theorem formula:

a^2 + b^2 = c^2

(x - 20)^2 + (x - 40)^2 = x^2

x^2 - 40x + 400 + x^2 - 80x + 1600 = x^2

x^2 -120x + 2000 = 0

(x - 100)(x - 20) = 0

x - 100 = 0 or x - 20 = 0

x = 100 or x = 20

we see that the solution x = 20 must be discarded because it will give a negative side length and a side length of 0. the only valid solution is x = 100.

perimeter = sum of the lengths of the sides

perimeter = x + x - 20 + x - 40

perimeter = 3x - 60

replace x with 100.

perimeter = 3(100) - 60

perimeter = 300 - 60

perimeter = 240