Draw a line that is only in quadrants II III IV. The y-intercept is greater than or equal to -3 and less than or equal to -6 What id the line and the equation?

; )

step-by-step explanation: ; )

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

hello,

as we don't know if g(x) is continue,

answer b

f(x)=2x²-8x+1=2(x²-4x+4)+1-8=2(x-2)²-7

vertex is (2,-7) ( a minimum since coefficient of x² is > 0)

example answer: "to determine if a sum is reasonable you can round each number to the nearest whole and compare the estimated sum to the actual sum."

hope this you and anyone else who needs the answer : ^)