All categories

3 years ago

Help please ????????

Mathematics

2 answers:

julsineya [31]3 years ago

6 0

8+8+6+6+12+12

16+12+24

40+12

Naily [24]3 years ago

5 0

The surface area of this prism is equal

two areas of rectangular 2m × 4m + two areas of rectangular 2m × 3m + two areas of rectangular 3m × 4m

Calculate:

$SA=2\cdot2\cdot4+2\cdot2\cdot3+2\cdot3\cdot4=52 m^2$

Answer: 52 m²

You might be interested in

340,250 rounded to the nearest thousand?

kondor19780726 [428]

Answer:

340,000

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

3. The Parkers spent $142.78, $145.91, and $141.15 in 3 visits to the

vladimir2022 [97]

Answer:

$20.16

Step-by-step explanation:

all the costs from the trips added up is $429.84, so subtract that from $450.00 and that's the answer

4 0

3 years ago

X + 3 > 4 can anybody tell me how to solve this question today please

pav-90 [236]

Subtract 3 from both sides

x>-1 is the answer

8 0

3 years ago

Evan bought two plants. He decided to water his first plant every 3 days and his second plant every 4 days. If he watered both o

nata0808 [166]

Answer:

the 12th

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Now, lets evaluate the same integral using power series. first, find the power series for the function f(x = \frac{32}{x^2+4}. t

BlackZzzverrR [31]

No idea what the previous part of the problem is, but you have

$f(x)=\dfrac{32}{x^2+4}=\dfrac8{1-\left(-\frac{x^2}4\right)}=\displaystyle8\sum_{n\ge0}\left(-\frac{x^2}4\right)^n$
$f(x)=\displaystyle8\sum_{n\ge0}\left(-\dfrac14\right)^nx^{2n}$

which is valid for $\left|-\dfrac{x^2}4\right|$ , or $|x|$ . So the integral from 0 to 2 is

$\displaystyle\int_0^2f(x)\,\mathrm dx=\int_0^28\sum_{n\ge0}\left(-\frac14\right)^nx^{2n}\,\mathrm dx$
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\int_0^2x^{2n}\,\mathrm dx$

Note that since the power series only converges on the interval if $x$ is strictly less than 2, which means we have to treat this as an improper integral.

$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\int_0^tx^{2n}\,\mathrm dx$ [/tex]
$=\displaystyle8\sum_{n\ge0}\left(-\frac14\right)^n\lim_{t\to2^-}\frac{x^{2n+1}}{2n+1}\bigg|_{x=0}^{x=t}$
$=\displaystyle8\sum_{n\ge0}\frac{(-1)^n}{2^{2n}(2n+1)}\lim_{t\to2^-}t^{2n+1}$
$=\displaystyle16\sum_{n\ge0}\frac{(-1)^n}{2n+1}$
$=16-\dfrac{16}3+\dfrac{16}5-\dfrac{16}7+\dfrac{16}9+\cdots$

6 0

3 years ago