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SOMEONE PLEASE HELP find the greatest common factor 14g5h2, 56g3h4, 77g3h2

andrezito [222]

Answer:

$\large\boxed{GCF(14g^5h^2,\ 56g^3h^4,\ 77g^3h^2)=g^3h^2}$

Step-by-step explanation:

$14g^5h^2=2\cdot7\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot g\cdot g\cdot \boxed{h}\cdot \boxed{h}\\\\56g^3h^4=2\cdot2\cdot2\cdot7\cdot\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}\cdot h\cdot h\\\\77g^3h^2=7\cdot11\cdot\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}\\\\GCF(14g^5h^2,\ 56g^3h^4,\ 77g^3h^2)=\boxed{g}\cdot \boxed{g}\cdot \boxed{g}\cdot \boxed{h}\cdot \boxed{h}=g^3h^2$

8 0

3 years ago

(6x3 + 3x² + 3) + (2x - 5x + 1) the sum is?

laiz [17]

<h2>Answer:</h2><h2>8x^3 + 3x^2 - 5x + 4 is the answer.</h2>

Step-by-step explanation:

(6x^3 + 3x² + 3) + (2x^3 - 5x + 1)

6x^3 + 2x^3 = 8x^3

3x^2 + 0 = 3x^2

-5x + 0 = -5x

3 + 1 = 4

8x^3 + 3x^2 - 5x + 4 is the answer because you have to put all the terms that were solved together. This will lead to 8x^3 + 3x^2 - 5x + 4. And, standard form is ax^2 + bx + c. So, 8x^3 + 3x^2 - 5x + 4 is the answer.

8x^3 + 3x^2 - 5x + 4

Hope this helped,

Kavitha

8 0

3 years ago

Read 2 more answers

A system of equations is given below:

DaniilM [7]

You have the value for y, so you substitute that equation for y, times three because there are three yS.

So you have: 2x + 3y = 10

2x + 3 x 2x + 4 or 2x + 3(3x + 4)

so the answer would be B, 2x + 3(3x + 4)

4 0

3 years ago

What are the numbers that have an absolute value of 2 and 1 half

rewona [7]

Absolute value of |2 1/2| is 2 1/2

4 0

3 years ago

Read 2 more answers

Using the Factor Theorem, which of the polynomial functions has the zeros 2, radical 3 , and negative radical 3 ? f (x) = x3 – 2

Basile [38]

Answer:

A

$f(x) = x^3 - 2x^2 -3x + 6$

Step-by-step explanation:

According to the Factor Theorem, if (x - k) is a factor of a polynomial P(x), then P(k) must equal zero.

We are given that a polynomial function has the zeros 2, √3, and -√3. So, we can let k = 2, √3, -√3.

So, according to the Factor Theorem, P(2), P(√3) and P(-√3) must equal 0.

Testing each choice, we can see that only A is true:

$\displaystyle f(x) = x^3 - 2x^2 - 3x + 6$

Testing all three values yields that:

$\displaystyle \begin{aligned} f(2) &= (2)^3 - 2(2)^2 -3(2) + 6 \\ &= (8) - (8) -(6) + (6) \\ &= 0\stackrel{\checkmark}{=}0 \\ \displaystyle f(\sqrt{3}) &= (\sqrt{3})^3 - 2(\sqrt{3})^2 - 3(\sqrt{3}) + 6 \\ &=(3\sqrt{3}) -(6)-(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \\ f(-\sqrt{3}) &= (-\sqrt{3})^3 - 2(-\sqrt{3})^2 - 3(-\sqrt{3}) + 6 \\ &=(-3\sqrt{3}) -(6)+(3\sqrt{3}) + 6 \\ &= 0\stackrel{\checkmark}{=}0 \end{aligned}$

Hence, our answer is A.

3 0

2 years ago