All categories

3 years ago

Brainliest! Brainliest! Brainliest!

Mathematics

2 answers:

emmasim [6.3K]3 years ago

6 0

Angle F because its rays form straight lines with the rays of angle h, which is why they are vertical.

SIZIF [17.4K]3 years ago

3 0

I believe it’s angle f

You might be interested in

Points a, b, c are collinear. Point b is between A and C. AC=3x+3, ab=-1+2x, and BC=11. solve x.

romanna [79]

X is 7. Hope this helps! Photo above

5 0

3 years ago

The Oliver Company plans to market a new product. Based on its market studies, Oliver estimates that it can sell up to 5,500 uni

Ghella [55]

For the Oliver Company to break even, the total revenue must equal the sum of the variable costs and the fixed cost. Mathematically, this can be represented as: Total revenue = 0.4*(Total revenue) + (Fixed Costs) Let the number of units sold be x. then, 7*x = 0.4*(7*x) + 6300 Thus, x = 6300/(0.6*7) = 1500 units. Thus the company will have to sell 1500 units to break even.

8 0

3 years ago

If a horizontal line segment is 5 units in length,What is the endpoint for the line segment?

IceJOKER [234]

Answer: -3,7

Step-by-step explanation:

4 0

4 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.

5 0

3 years ago

The image of (6, 9) under a dilation is (4, 6).

Ad libitum [116K]

For this case we have the following ordered pairs:
(6, 9)
(4, 6)
The scale factor can be found in two different ways:
$k = 6/4 = 3/2 = 1.5

k = 9/6 = 3/2 = 1.5$
Therefore, the scale factor is given by:
$k = 1.5
$
Answer:
The scale factor of the dilation is 1.5

8 0

3 years ago

Read 2 more answers