Find 3/7 of 1/5 please helopp

8/35

Step-by-step explanation:

Step 1

We can't subtract two fractions with different denominators. So you need to get a common denominator. To do this, you'll multiply the denominators times each other... but the numerators have to change, too. They get multiplied by the other term's denominator.

So we multiply 3 by 5, and get 15.

Then we multiply 1 by 7, and get 7.

Next we give both terms new denominators -- 7 × 5 = 35.

So now our fractions look like this:

15

35

−

7

35

Step 2

Since our denominators match, we can subtract the numerators.

15 − 7 = 8

So the answer is:

8

35

Step 3

Last of all, we need to simplify the fraction, if possible. Can it be reduced to a simpler fraction?

To find out, we try dividing it by 2...

Nope! So now we try the next greatest prime number, 3...

Nope! So now we try the next greatest prime number, 5...

Nope! So now we try the next greatest prime number, 7...

Nope! So now we try the next greatest prime number, 11...

No good. 11 is larger than 8. So we're done reducing.

There you have it! The final answer is:

3/7

−

1/5

=

8/35

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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