All categories

3 years ago

A triangle with vertices at A(-2, 2), B(-2, 5), and C(-4, 5) is TRANSLATED. The image of vertex A has coordinates at (4, -3)

Mathematics

1 answer:

scZoUnD [109]3 years ago

3 0

Step-by-step explanation:

so sorry

don't know but please mark me as brainliest please

You might be interested in

Hamid bought 4.5 pounds of strawberries. The total cost of the strawberries was 8.10. How much did each pound of strawberries co

Ray Of Light [21]

Answer:

<h2>1.80 per pound of strawberries</h2><h2 />

Step-by-step explanation:

8.10 / 4.5 pounds = 1.80 per pound of strawberries

4 0

3 years ago

Read 2 more answers

Does anyone add me on hangout boys and girls anyone myself sahil from jharkhand class 9

bazaltina [42]

Answer: Hm, this sounds a bit out of order I hope you use better grammar and please ask a question not a comment

4 0

3 years ago

Rewrite as a simplified fraction. 2.3 = ?

zubka84 [21]

Answer:

23/100 or 23/10

Step-by-step explanation:

2.3/1

(2.3 x 10) / (1 x 10) =23/10

find lcm (lowest common multiple) for 23 and 10

1 is the lcm for 23 and 10

23/10 is a simplest fraction for the decimal point number 2.3

7 0

3 years ago

What is the value of the 5 in the number 356,409?

Hoochie [10]

Answer:

ten thousands

Step-by-step explanation:

THIS IS MIDDLE SCHOOL MATH IM SORRY

3 0

3 years ago

Read 2 more answers

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