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3 years ago

Which equation represents the line that passes through (0, 2) and (3, 1)?

Mathematics

1 answer:

Anna11 [10]3 years ago

6 0

<h3>Answer:</h3>

A. y = -1/3x + 2 . . . . . . . . assuming a typo in your problem statement

<h3>Explanation:</h3>

The point (0, 2) is the y-intercept. This tells you the equation will be something of the form ...

... y = mx + 2 . . . . . . matching only answer choices A and C

_____

Knowing this, you can either substitute the values x=3, y=1 into these two equations to see which one works (it is not C), or you can look at the two points and determine the value of the slope.

The slope is ...

... (difference in y)/(difference in x) = (1-2)/(3-0) = -1/3 . . . . . matches choice A

The equation of the line is ...

... y = -1/3x + 2 . . . . . matches answer choice A

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Answer:

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3 years ago

Through: (4,2), slope = 1/4

sattari [20]

Answer:

y = 1/4x + 1

Step-by-step explanation:

y = 1/4 x + b

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3 years ago

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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