Melanie buys 6 pounds of beef and 3 pounds of bacon. She pays a total of $63, and the bacon costs $1.50 less per pound than the beef. What would be the
combined cost of 1 pound of beef and 1 pound of bacon? *

Cost of 1 beef= $7.5

Cost of 1 bacon= $6

Step-by-step explanation:

Ok so first of organize everything:

We know she buys 6 pounds ->beef and 3 pounds ->bacon

First we need to set up an equation of how much 1 beef costs and 1 bacon costs:

Let’s say f=cost of 1 beef and b=cost of 1 bacon

Now lets make the equation:

6f+3b=63. (She buys 6 portions of beef and 3 bacon and her total comes out to 63)

Now their is another condition that the cost of one bacon is -$1.50 per pound than beef.

So we can say b=f-1.50

Do you notice anything similar? The b has a value now and so we can replace the b in our first equation with the b of the 2nd (b=b)

So hence,

6f+3(f-1.50)=63

6f+3f-4.5=63

9f=67.5

F=7.5 (This is the cost for 1 pound of beef)

Now lets find out for bacon,. Now instead of doing this equation all over again just insert it into the b=f-1.5 you know make life easier for yourself:

B=7.5-1.5

B=6

Cost of 1 beef= $7.5

Cost of 1 bacon= $6

ce pula mea vrei ma

step-by-step explanation:

gayule

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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