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prisoha [69]
3 years ago
9

Hello, happy Friday, I am just here with some geometry questions.

Mathematics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

  • U'(-14, 8)

Step-by-step explanation:

<u>The transformations include:</u>

  • T<-2,2>·D3

This is a dilation by a scale factor of 3 and then translation 2 units left and 2 units up.

<u>The transformation applied to the point U:</u>

  • U(-4,2) → U'(-4*3 - 2, 2*3 + 2) = U'(-14, 8)
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Find maclaurin series
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Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

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2 years ago
Kristi is the food manager of a summer camp. Each day, the kids at the camp get a choice of a pizza, chicken nuggets, fish, or s
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Kristi's prediction is not valid. The probability of a kid choosing a fish plate is about 21%.

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I took the test ✌

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3 years ago
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let's say is 16% of "x", meaning "x" is the 100%.

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