Approximate -√23 to the nearest tenth

The first line in a system of linear equations has a slope of 3 and passes through the point (-1 , -8). The second line passes t

gregori [183]

The equation of the first line can be written in point-slope form as
.. y = 3(x +1) -8
or
.. 3x -y = 5

The equation of the second line can be written in 2-point form as
.. y = (-1-3)/(10-(-6))*(x +6) +3
.. y = (-1/4)(x +6) +3
or
.. x +4y = 6

A graph shows the solution to this system is (x, y) = (2, 1).

_____
The second equation can be used to write an expression for x:
.. x = 6 -4y
This can be substituted into the first equation.
.. 3(6 -4y) -y = 5
.. 18 -13y = 5 . . . . . . . collect terms
.. 13 = 13y . . . . . . . . . add 13y-5
.. 1 = y . . . . . . . . . . . . divide by 13
From the above equation for x
.. x = 6 -4*1 = 2

7 0

4 years ago

How much sit-ups can you do in 1 minutes?

galben [10]

I can do like 42, my highest was 54 though

8 0

3 years ago

there are 25 kids who are playing with a parachute, but 45 kids in total can play with the parachute, how many more kids can joi

Fynjy0 [20]

20. 25 + 20 = 45. hope this helps :)

3 0

3 years ago

Read 2 more answers

(2,-4), (x,\ y), A, and B all lie on the line. Find an equation relating x and y.

Olin [163]

Answer:

I can't see the picture, sorry. :(

Step-by-step explanation:

4 0

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,

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

(c) The series

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

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.

5 0

3 years ago