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inessss [21]
3 years ago
15

The archway of the main entrance of a university is modeled by the quadratic equation y = -x2 + 6x. The university is hanging a

banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?

Mathematics
1 answer:
Yuliya22 [10]3 years ago
6 0

Answer:

The point at which the banner should be attached to the archway is going to be given by the point(s) at which the two equations intercept:

y = -x^2 + 6x and 4y = 21 − x

Solving the system of equations, we find that they intercept at: (1, 5) and (5.25, 3.938)

Therefore, the banner should be attached to the archway at the points (1, 5) and (5.25, 3.938)

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Answer:

We have the sentence:

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Let's break it into parts.

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This can be written as:

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"...  2 by the power of -2 times x by the power of 0times x by the power of 9"

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\frac{x^5*y^6}{2^{-2}*x^0*x^9}

And any number by the power of 0 is equal to 1, then:

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\frac{x^5*y^6}{2^{-2}*x^9}

We can keep simplifying this.

We know that:

a^(-n) = (1/a)^(n)

Then:

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Then we get:

\frac{x^5*y^6}{2^{-2}*x^9} = \frac{x^5*y^6}{x^9}*4

And we also know that:

a^n/a^m = a^(n - m)

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And we can't simplify this anymore.

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