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3 years ago

F(x)=2(x+1)^2-2 What is the vertex

Mathematics

1 answer:

Bumek [7]3 years ago

8 0

Answer:

(-1,-2)

Step-by-step explanation:

Let's bring back to memory vertex formula for parabolas.

$y=a(x-h)^{2}+k$

Where (h,k) is the vertex.

In $f(x)=2(x+1)^{2}-2$

The vertex is (-1,-2)

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Let f be defined by the function f(x) = 1/(x^2+9)

riadik2000 [5.3K]

(a)

$\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}$

Substitute x = 3 tan(t ) and dx = 3 sec²(t ) dt :

$\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt$

$=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}$

(b) The series

$\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}$

converges by comparison to the convergent p-series,

$\displaystyle\sum_{n=3}^\infty\frac1{n^2}$

$\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}$

converges absolutely, since

$\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}$

That is, ∑ (-1)ⁿ (n ² + 9)/eⁿ converges absolutely because ∑ |(-1)ⁿ (n ² + 9)/eⁿ| = ∑ (n ² + 9)/eⁿ in turn converges by comparison to a geometric series.

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3 years ago

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Gekata [30.6K]

Answer:

Step-by-step explanation:

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VashaNatasha [74]

Answer:

9 ft

Step-by-step explanation:

You can use cosine to solve this, which is adjacent over hypotenuse. The ladder forms a triangle with the house, and the side you need to solve for is adjacent to the angle. You can set up the equation:

cos(50)=x/14

Then you multiply 14 by cos(50) and get almost 9 ft (rounding to the nearest tenth of a foot).

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4 years ago

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viva [34]

Answer:

-1/2x+1 is the equation.

Step-by-step explanation:

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