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2 years ago

Triangle ABC has an area of 25 square feet and a perimeter of 65.5 feet. △ABC is dilated by a factor of 5/2 to create △DEF. \

Mathematics

1 answer:

VARVARA [1.3K]2 years ago

8 0

Answer:

Solution. To find the perimeter of \triangle ABC we use the Pythagorean Theorem which tells us that |AB|^2 = |AC|^2 + |BC|^2. Since |AC| = 5 and |BC| = 12 we find that |AB| = \sqrt{169} = 13. The perimeter of \triangle ABC is then 5 + 12 + 13 = 30 units.

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Answer:

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Step-by-step explanation:

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4 years ago

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

Pavel [41]

Answer:

The critical numbers/values are x = 0, 4/7, 2

Step-by-step explanation:

This is a doozy; no wonder you have it up here for help!

The critical numbers of a function are found where the derivative of the function is equal to 0. To find these numbers, you have to factor the deriative or simply solve it for 0. This one is especially difficult since it involves rational exponents that have to be factored. But this is fun, so let's get to it.

First off, I am assuming that the function is

$f(x)=x^{\frac{4}{5}}*(x-2)^2$ which involves using the product rule to find the derivative.

That derivative is

$f'(x)=x^{\frac{4}{5}}*2(x-2)+\frac{4}{5}x^{-\frac{1}{5}}(x-2)^2$ which simplifies down to

$f'(x)=x^{\frac{4}{5}}(2x^{\frac{5}{5}}-4)+\frac{4}{5}x^{-\frac{1}{5}}(x^{\frac{10}{5}}-4x^{\frac{5}{5}}+4)$ and

$f'(x)=2x^{\frac{9}{5}}-4x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}$

Let's get everything over the common denominator of 5 so we can easily add and subtract like terms:

$f'(x)=\frac{10}{5}x^{\frac{9}{5}}-\frac{20}{5}x^{\frac{4}{5}}+\frac{4}{5}x^{\frac{9}{5}}-\frac{16}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}$

Combining like terms gives us

$f'(x)=\frac{14}{5}x^{\frac{9}{5}}-\frac{36}{5}x^{\frac{4}{5}}+\frac{16}{5}x^{-\frac{1}{5}}$

This, however, factors so it is easier to solve for x. First we will set this equal to 0, then we will factor out

$\frac{2}{5}x^{-\frac{1}{5}}$ :

$0=\frac{2}{5}x^{-\frac{1}{5}}(7x^2-18x+8)$

By the Zero Product Property, one of those terms has to equal 0 for the whole product to equal 0. So

$\frac{2}{5}x^{-\frac{1}{5}}=0$ when x = 0

And

$7x^2-18x+8=0$ when x = 2 and x = 4/7

Those are the critical numbers/values for that function. This indicates where there is a max value or a min value.

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