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____ [38]
2 years ago
7

Solve the following: -5² + 10² (2 × (-5) × 3) + 3³ Give your answer in simplest form.

Mathematics
1 answer:
notsponge [240]2 years ago
5 0
Problem 1: -5^2 + 10^2
Work:
-5^2 + 10^2—> (-5 x -5) + (10 x 10) —> 25 + 100 = 125
Answer:
125

Problem 2: (2 x (-5) x 3) + 3^3
Work:
(2 x (-5) x 3) + 3^3 > (-25 x 3) + 3^3 —> -75 + 3^3 —> -75 + 27 —> 48
Answer:
48
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Korolek [52]

Answer:

0.01893491124

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If you meant 845 divided by 16 then the answer is 52.8125.

Please tell me if this helped you out

4 0
2 years ago
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Sophie [7]
The answer is C if not sorry
7 0
3 years ago
Find maclaurin series
Mumz [18]

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}

7 0
2 years ago
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PIT_PIT [208]

Answer:

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Cost per bagel = $0.60

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2 years ago
!NO LINKS, NO FILES!<br><br> please :D
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Answer:

I believe it's 12

Step-by-step explanation:

I am so sorry if it's incorrect-

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3 years ago
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