Al saves pennines. he agreed to give 6/13 of his pennines to bev if she would give 6/13 of what she got from al to carl and if carl in turn would give 6/13 of what he got from bev to dani. bev, carl, and dani agreed and dani received 2376 pennines. how many pennies did al have initially,

24,167 pennies

Step-by-step explanation:

Let's see what happens if we say Al had 1000 pennies.

Bev would get 6/13 of 1000 which is  462 pennies. ((6/13)*1000)

Continuing like this Carl would get (6/13)*462 = 213

Dani would get 213(6/13)=98

Now, another way of writing this is 1000*(6/13)*(6/13)*(6/13).  hopefully that makes sense, if not let me know.

Anyway, what is another way of writing this?  If it isn't clear how else can you write 3*3*3*3?  Exponents is the answer.  3*3*3*3=3^4 and (6/13)*(6/13)*(6/13)= (6/13)^3, so you can write the whole equation as 1000(6/13)^3=98

Now, using the information given in the question we know how much is gotten by Dani as well as (6/13)^3, but we don't know how much we start with.  So instead of 1,000 like I showed lets use a variable x.

x*(6/13)^3 = 2376

Now you can use algebra to solve for x

x*(6/13)^3 = 2376 step one let's just figure out what (6/13)^3 is.

x*(216/2197) = 2376  Now here you can either divide by 216/2197 or multiply by 2197 then divide by 216, it's the same thing.

x = 24167

You can double check too with 24167*(6/13)^3

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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