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leva [86]
3 years ago
7

Your teacher recently found an old Nintendo in a thrift store. Use the formula to predict the value of the Nintendo in 2020. (Re

member, x is the years after 1986, and y is the
value, in $.) Do you think this is a realistic prediction of the value of that Nintendo?
Formula- f(x) = 3x^2 - 40x + 180
Mathematics
1 answer:
9966 [12]3 years ago
4 0

The value of the Nintendo after 35 years is $2455

Since the formula f(x) = 3x² - 40x + 180 predicts the value of the Nintendo x years after 1986.

Since we require the value in 2021, x years after 1986 is 2021 - 1986 = 35 years.

Substituting x = 35 into the equation, we have

f(x) = 3x² - 40x + 180

f(x) = 3(35)² - 40(35) + 180

f(x) = 3(1225) - 40(35) + 180

f(x) = 3675 - 1400 + 180

f(x) = 2275 + 180

f(x) = 2455

So, the value of the Nintendo after 35 years is $2455

Do you think this is a realistic prediction of the value of that Nintendo?

This is not a realistic prediction for the value of the Nintendo, because, it is too high.

Learn more about quadratic equations here:

brainly.com/question/13704125

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Your answer is 725 hope this helps 
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4 years ago
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AleksAgata [21]

Answer:

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Step-by-step explanation:

Let's start by breaking down each of the radicals:

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Since we're dealing with a cube root, we'd like to pull as many perfect cubes out of the terms inside the radical as we can. We already have one obvious cube in the form of x^3, and we can break 16 into the product 8 · 2. Since 8 is a cube root -- 2³, to be specific, we can reduce it down as we simplify the expression. Here our our steps then:

\sqrt[3]{16x^3y}\\=\sqrt[3]{2\cdot8\cdot x^3\cdot y}\\=\sqrt[3]{2} \sqrt[3]{8} \sqrt[3]{x^3} \sqrt[3]{y} \\=\sqrt[3]{2} \cdot2x\cdot \sqrt[3]{y} \\=2x\sqrt[3]{2}\sqrt[3]{y}

We can apply this same technique of "extracting cubes" to the second term:

\sqrt[3]{54x^6y^5} \\=\sqrt[3]{2\cdot27\cdot (x^2)^3\cdot y^3\cdot y^2} \\=\sqrt[3]{2}\sqrt[3]{27} \sqrt[3]{(x^2)^3} \sqrt[3]{y^3} \sqrt[3]{y^2}\\=\sqrt[3]{2}\cdot 3\cdot x^2\cdot y \cdot \sqrt[3]{y^2} \\=3x^2y\sqrt[3]{2} \sqrt[3]{y}

Replacing those two expressions in the parentheses leaves us with this monster:

2(2x\sqrt[3]{2}\sqrt[3]{y})+4(3x^2y\sqrt[3]{2} \sqrt[3]{y})

What can we do with this? It seems the only sensible thing is to look for terms to factor out, so let's do that. Both terms have the following factors in common:

4, \sqrt[3]{2} , x

We can factor those out to give us a final, simplified expression:

4\sqrt[3]{2}x(\sqrt[3]{y}+3xy\sqrt[3]{y} )

Not that this is the same sum as we had at the beginning; we've just extracted all of the cube roots that we could in order to rewrite it in a slightly cleaner form.

6 0
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Answer:

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Step-by-step explanation:

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