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oee [108]
3 years ago
13

A club sold $555 worth of

Mathematics
2 answers:
Ivan3 years ago
7 0

Answer:

111 tickets were sold.

Step-by-step explanation:

If each ticket costs 5$, and they made 555$, then by dividing 555 by 5, you can find 111 tickets as your answer.

anastassius [24]3 years ago
6 0

Answer:

111

Step-by-step explanation:

$555 ÷ $5 = 111

sorry I'm not good at explaining

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Five people are at a party. Everyone shakes hands once with everyone else. How many handshakes are there? Show your
blagie [28]

Answer:

10

Step-by-step explanation:

1 2 3 4 5

1 shakes hands with 2,3,4,5

2 shakes hands with 3,4,5 but not with 1 because they already shook hands once

similarly, 3 only shakes hands with 4 and 5, same reason as abov

4 with 5 only, same reason

7 0
3 years ago
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WHOEVER ANSWERS CORRECTLY W STEP BY STEP GETS BRAINLIEST
ANTONII [103]

Answer:

313 in.

Step-by-step explanation:

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2 years ago
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(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
3 years ago
Which factoring technique could you use to factor the following polynomial:
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The ans shld be b blah blah
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Put in order from large to small:<br><br> 0,2,-1.5,-2,0.5,-5
Dafna1 [17]
0.5, 0.2, -1.5 , -2 , -5
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