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Over [174]
3 years ago
11

Find the product. (0.5n^5)^2(10n^7)^3

Mathematics
2 answers:
pochemuha3 years ago
5 0
The answer is 250x^31

never [62]3 years ago
3 0

Answer:

250{n}^{31}

Step-by-step explanation:

(0.5 {n}^{5} )^{2} (10 {n}^{7} )^{3}  \\  = (0.5)^{2}{n}^{5 \times 2} \times (10)^{3} {n}^{7 \times 3} \\  = 0.25{n}^{10} \times 1000{n}^{21} \\  = 0.25 \times 1000 {n}^{10 + 21} \\  = 250{n}^{31}

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Let n be a whole number, and consider the statements below. p: n is a multiple of two. q: n is an even number. Which of the foll
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A farmer sells 8.8 kilograms of apples and pears at the farmer's market. 3 /4 of this weight is apples, and the rest is pears. H
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3 years ago
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The roots of an equation are simply the x-intercepts of the equation.

See below for the proof that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

The equation is given as: \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

There are several ways to show that an equation has real roots, one of these ways is by using graphs.

See attachment for the graph of \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0}

Next, we count the x-intercepts of the graph (i.e. the points where the equation crosses the x-axis)

From the attached graph, we can see that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} crosses the x-axis at approximately <em>-2000 and 2000 </em>between the domain -2500 and 2500

This means that \mathbf{\frac{x^4}{2021} = 2021x^2 - x - 3 = 0} has at least two real roots

Read more about roots of an equation at:

brainly.com/question/12912962

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3 years ago
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