For what nonzero whole number k does the quadratic equation y^2+2ky+4k=0 have exactly one real solution for y?

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

answer:

c) the x-coordinate of the center of the circle increases by 3

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5