. Consider this word problem about baking brownies: You are baking brownies for your class. You put white frosting on 1 you put small red hearts on 1 3 of the brownies and 4 of the brownies. How many brownies have both white frosting and small red hearts on them?

The fraction of brownies with both white frosting and small red hearts is not given. And also the total number of brownies made is not given too.

Therefore, the number of brownies that have both white frosting and small red hearts on them cannot be determined

Step-by-step explanation:

Given;

1. You put 1 white frosting per 3 of the brownies

2. You put 1 small red hearts per 4 of the brownies

Fraction of brownies with;

White frosting = 1/3

Small red hearts = 1/4

The fraction of brownies with both white frosting and small red hearts is not given. And also the total number of brownies made is not given too.

Therefore, the number of brownies that have both white frosting and small red hearts on them cannot be determined

For example:

Assuming that;

The fraction of brownies with both white frosting and small red hearts = 1/12

The total number of brownies baked = 36

The number of brownies with both white frosting and small red hearts = 1/12 × 36 = 3 brownies.

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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