*PLEASE HELP ME* (math attachment) IT'S PAST MIDNIGHT I'M TIRED AND DESPERATE

Step-by-step explanation:

Z id for fun

9038735228

pwd -97cEeT

only for Interested g.i.r.l.s

D

let's rewrite this

x + y = 11                  - equation 1

   - x = - y - 9           - equation 2

solve the first equation for x

x + y = 11            (add -y to both sides to get x by itself)

     x = -y + 11           - equation 3

now substitute -y + 11 for x in the second equation

 y - 11 = - y - 9        (2 negative signs next to each other equals positive)

2y - 11 = -9              (add y to both sides)

      2y = 2               (add 11 to both sides)

        y = 1                (divide both sides by 2)

now substitute y = 1 into equation 3

x = -1 + 11

x = 10

therefore, x = 10 and y = 1. in coordinates, x goes before y

therefore, the coordinates are (10, 1)

therefore the answer is D

good luck my friend! hope this helped a lot! also, i found a website if you don't fully understand this, its called MathPapa, i didn't use it to work this out so my working might be different but it definitely helps me sometimes. for an equation like this, type it likes this: x+y=11; -x=-y-9

-cheesetoasty

mxb67

step-by-step explanation:

answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

step-by-step explanation:

polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

you 're wlecome! : )

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