Ralph has bought a variety of 35 donuts. one type costed $2.35 per donut while the other costed 1.75 per donut. he spent $69.05. how many of each donut did ralph buy?

He bought 22 of one type and 13 of the other type of donut.

Step-by-step explanation: Since the question stated that he bought two types of donuts, we would call the donuts type a and type b. If he bought a variety of 35 donuts, then we can express this as;

a + b = 35

Also, if one type cost $2.35 and another type cost $1.75 and the total cost was $69.05, this we can also express as,

2.35a + 1.75b = 69.05

We now have a pair of simultaneous equations as follows;

a + b = 35 ———(1)

2.35a + 1.75b =69.05 ———(2)

From equation (1), we make a the subject of the equation, a = 35 - b

Substitute for the value of a into equation (2)

2.35(35 - b) + 1.75b = 69.05

82.25 - 2.35b + 1.75b = 69.05

Collect like terms and we have

83.25 - 69.05 = 2.35b - 1.75b

13.2 = 0.6b

Divide both sides of the equation by 0.6

22 = b

With the value of b now known, substitute for the value of b into equation (1)

a + b = 35

a + 22 = 35

Subtract 22 from both sides of the equation

a = 13.

Therefore, Ralph bought 22 of one type, and 13 of the other type of donut

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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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