PPPPLLLZZZ HHHEELLPPP!!! BRAINLIESTTTT AND 15 POINTZ!!!!

Mathematics

2 answers:

coldgirl [10]3 years ago

8 0

Yeah, I’d go with 1,555 as well.

Hope this helps!! ;))
Have a grat day!! <3

Savatey [412]3 years ago

6 0

I would say around 1555 people would like polar bears.

You might be interested in

The members of two running clubs entered their daily running totals into a computer program. The leaders of the two clubs random

Delvig [45]

C should be the right answer

6 0

3 years ago

Solve for L: P = 2L + 2W

natulia [17]

$P = 2L + 2W\\2L=P-2W\\L=\dfrac{P-2W}{2}$

3 0

3 years ago

(2,-2) parallel to y=-3/2x + 5

garri49 [273]

The answer to the question

5 0

4 years ago

Kenneth's Tea Shop has caffeinated tea and decaffeinated tea. The tea shop served 63 caffeinated

UNO [17]

Answer:

90%

90% of the teas were caffeinated.

Step-by-step explanation:

Total teas: 63+7= 70

63/70 x 100/1= 90%

90% of the teas were caffeinated.

4 0

3 years ago

(a) Use the reduction formula to show that integral from 0 to pi/2 of sin(x)^ndx is (n-1)/n * integral from 0 to pi/2 of sin(x)^

Sedbober [7]

Hello,

a)
$I= \int\limits^{ \frac{\pi}{2} }_0 {sin^n(x)} \, dx = \int\limits^{ \frac{\pi}{2} }_0 {sin(x)*sin^{n-1}(x)} \, dx \\

= [-cos(x)*sin^{n-1}(x)]_0^ \frac{\pi}{2}+(n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos(x)*sin^{n-2}(x)*cos(x)} \, dx \\

=0 + (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {cos^2(x)*sin^{n-2}(x)} \, dx \\

= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {(1-sin^2(x))*sin^{n-2}(x)} \, dx \\
= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx - (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^n(x) \, dx\\

$
$I(1+n-1)= (n-1)*\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
I= \dfrac{n-1}{n} *\int\limits^{ \frac{\pi}{2} }_0 {sin^{n-2}(x)} \, dx \\
$

b)
$\int\limits^{ \frac{\pi}{2} }_0 {sin^{3}(x)} \, dx \\
= \frac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx \\
= \dfrac{2}{3}\ [-cos(x)]_0^{\frac{\pi}{2}}=\dfrac{2}{3} \\




$

$\int\limits^{ \frac{\pi}{2} }_0 {sin^{5}(x)} \, dx \\
= \dfrac{4}{5}*\dfrac{2}{3} \int\limits^{ \frac{\pi}{2} }_0 {sin(x)} \, dx = \dfrac{8}{15}\\





$

c)

$I_n= \dfrac{n-1}{n} * I_{n-2} \\

I_{2n+1}= \dfrac{2n+1-1}{2n+1} * I_{2n+1-2} \\
= \dfrac{2n}{2n+1} * I_{2n-1} \\
= \dfrac{(2n)*(2n-2)}{(2n+1)(2n-1)} * I_{2n-3} \\
= \dfrac{(2n)*(2n-2)*...*2}{(2n+1)(2n-1)*...*3} * I_{1} \\\\

I_1=1\\

$

3 0

4 years ago