A grocery store offers a cheese sampler that includes a pepper cheddar cheese that costs $12 per kilogram and Pennsylvania Jack that costs $11 per kilogram. How many kilograms of each were used to make a 5-kilogram mixture that costs $11.30 per kilogram?

3.5 kg of Pennsylvania Jack and 1.5 kg of Cheddar

Step-by-step explanation:

Let C be the kg of cheddar cheese and P be the kg of Pennsylvania Jack cheese needed for the sampler.

We know that C is $12/kg and P is $11/kg.

We can say that the total cost for a cheese sampler would be:

                 12C + 11P = Total Cost

We need to find a 5 kg mixture:  

                      C + P = 5 kg

that costs $ll.30/kg.  The total cost would be:  (5kg)($11.30/kg) = $56.50

                       C + P = 5 kg

Rearrange  C + P = 5 kg to:                      

                         C = 5kg - P and then use this in the cost equation:

                 12C + 11P = $56.50

                 12(5 - P) + 11P = 56.50

                  60 - 12P + 11P = 56.50

                       -P = - 3.50

                        P = 3.50 kg

That means we have:

                           C + P = 5 kg

                           C + 3.50 = 5 kg

                           C =  1.50 kg

Check to see if these values work:

                               12C + 11P = $56.50

                               12(1.50) + 11(3.50) = 56.50?

                                 18.00     + 38.50   =  56.50   YES

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! : )

cretdit to source

you didn't give me the choices, so i will just give you the graph so you can look at your choices and choose the right one.