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azamat
3 years ago
14

What is the relationship between Za and Zb?

Mathematics
1 answer:
Aloiza [94]3 years ago
8 0
They are complementary angles.
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Your sweet new bicycle has 2 front sprockets and 11 rear sprockets. Each distinct combination of a front sprocket with a rear sp
Naddika [18.5K]

Answer:

Your bicycle has 22 gears.

Step-by-step explanation:

The number of gears in the bycicle is the number of combinatiors of a front sprocket and a rear sprocket.

For each front sprocket, there are 11 rear sprockets. There are 2 front sprocts.

So, the number of gears that your bicycle has is

2*11 = 22.

8 0
3 years ago
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=1/x^5, y=0,
igomit [66]
If you wanted to rotate f(x) around the y axis from x=a to x=b then the volume would be
\pi  \int\limits^a_b {(f(x))^2)} \, dx


that would be

\pi \int\limits^8_1 {(\frac{1}{x^5})^2} \, dx
\pi \int\limits^8_1 {\frac{1}{x^{10}}} \, dx
1/x^10=x^-10
integrate
-1/(9x^9)
so
\pi \int\limits^8_1 {\frac{1}{x^{10}}} \, dx=
\pi [\frac{-1}{9x^{9}}]^8_1=\pi (\frac{-1}{9(8^{9})}-\frac{-1}{9(1^{9})})=
\pi (\frac{-1}{9(8^{9})}+\frac{1}{9(1^{9})})=\pi (\frac{-1}{9(8^{9})}+\frac{1}{9})=
\pi (\frac{-1}{9(8^{9})}+\frac{8^9}{9(8^9)})=\pi (\frac{8^9-1}{9(8^{9})})=
(\frac{\pi8^9-\pi}{9(8^{9})})=\frac{134217727 \pi}{1207959552}
that's the volume of the solid
4 0
4 years ago
A town’s population increases at a rate of 2.3% every year. The current population is 7,500 people. Which equation models this s
ruslelena [56]

Answer:

7500 ( 1.023 ) ^(x)

Step-by-step explanation:

x is the number of years that go by

3 0
3 years ago
Read 2 more answers
Hayden and his friends want to draw a court on the playground to play four square . The sides of each of the small square are 4
aksik [14]

Answer:

32

Step-by-step explanation:

let me know if that was correct

3 0
3 years ago
What’s is the sum of the series?
STALIN [3.7K]

Answer:

<h2>44</h2>

Step-by-step explanation:

\sum\limits_{k=1}^4(2k^2-4)\to a_k=2k^2-4\\\\for\ k=1:\\\\a_1=2(1)^2-4=2(1)-4=2-4=-2\\\\for\ k=2:\\\\a_2=2(2)^2-4=2(4)-4=8-4=4\\\\for\ k=3:\\\\a_3=2(3)^2-4=2(9)-4=18-4=14\\\\for\ k=4:\\\\a_4=2(4)^2-4=2(16)-4=32-4=28\\\\\sum\limits_{k=1}^4(2k^2-4)=-2+4+14+28=44

7 0
3 years ago
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