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3 years ago

What is the greatest common factor of 28x²y-7xy^5?

2 answers:

8 0

7xy would be the greatest common factor. You can divide both terms by 7xy. 7 goes into 28 4 times ad you can factor out both an x and a y from both terms. You would end up with: 7xy(4x-y^4).

Alchen [17]3 years ago

3 0

The greatest common factor of this can be solved by looking at the individual parts and splitting it up.

First, we have 28 and 7. Well, thats an easy one. 7 goes into 28 4 times so we are now left with 4 and 1.

We can also write the rest of this like this 4(x*x*y) - 1(x*y*y*y*y*y)

Now, what values are in both equations. We have one x and one y that can be taken out of both.

We end up with 7xy(4x-7y^4)

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2 years ago

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3 years ago

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neonofarm [45]

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6 0

3 years ago

Read 2 more answers

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Alex17521 [72]

Answer:

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(look at explanation)

Step-by-step explanation:

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3 years ago

1. Type an equation in the equation editor that uses 2 fractions with parentheses around one of them. Example: <img src="https:/

avanturin [10]

Answer:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$ $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii) $4^{3} + 8^{2} + \sqrt{9}$ $\Rightarrow$ 4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$ (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

Step-by-step explanation:

i) $\frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}$ $\Rightarrow$ \frac{3}{5} + (- \frac{1}{2}) = \frac{6}{10} - \frac{5}{10} = \frac{1}{10}

ii) $4^{3} + 8^{2} + \sqrt{9}$ $\Rightarrow$ 4^{3} + 8^{2} + \sqrt{9}

iii) $(\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}} \Rightarrow \hspace{0.2cm}$ (\frac{4}{5})^{2}. \sqrt[3]{8} \leqx^{3} - 3x + 6 \leq \sqrt{\frac{1}{3}}

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4 years ago