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GuDViN [60]
3 years ago
9

Evaluate the following expression. 6 squared -1

Mathematics
2 answers:
sesenic [268]3 years ago
7 0
The answer is 6^2=36-1= 35
natita [175]3 years ago
4 0

Answer:

35

Step-by-step explanation:

6^2 - 1\\36-1\\35

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Foci of ellipse of 3x2 + y2 =9
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Answer:

x=√-1/3 y^2 +3 or x=-√-1/3 y^2 +3

Step-by-step explanation:

Let's solve for x.

3x2+y2=9

Step 1: Add -y^2 to both sides.

3x2+y2+−y2=9+−y2

3x2=−y2+9

Step 2: Divide both sides by 3.

3x^2/3= -y^2+9/3

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Find maclaurin series
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\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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An incentive clause in Barry Bonds contract allows for a 3.4% bonus if he hits over 719 home runs in this season. His contract i
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