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2 years ago

A large beach ball has a diameter of 36 inches.what is the approximate volume of the beach ball?

2 answers:

8 0

Because a beach volleyball is a sphere, so the volume formula is V= 4pi r^3/r
so the answer would be 24,429.024

shtirl [24]2 years ago

7 0

V=24416.64 would be the volume of the entire beach ball

You might be interested in

How to solve all of this

photoshop1234 [79]

Start circle: πd = (3.14)(19) = 59.7

Move diagonally to the circle with the radius of 6.2.

Second circle: 2πr = 2(3.14)(6.2) = 39

Move upwards to the circle with the radius of 10.5

third circle: 2πr = 2(3.14)(10.5) = 66

Move right to the circle with the diameter of 16.6

Fourth circle: πd = (3.14)(16.6) = 52.2

Move down to the circle with the diameter of 7.7

fifth circle: πd = (3.14)(7.7) = 24.2

Move down to the circle with the diameter of 50

Sixth circle: πd = (3.14)(50) = 157.1

Move left to the circle with the radius of 11.8

Seventh circle: 2πr = 2(3.14)(11.8) = 74.1

Move down to the circle with the radius of 38

Eight circle: 2πr = 2(3.14)(38) = 238.8

Move right to the circle with the diameter of 1.1

ninth circle: πd = (3.14)(1.1) = 3.5

Move right to the circle with the radius of 14.8

10th circle = 2πr = 2(3.14)(14.8) = 93

Move up to the end.

Hope this helps :)

7 0

2 years ago

A particle moves according to a law of motion s = f(t), t ? 0, where t is measured in seconds and s in feet.

Usimov [2.4K]

Answer:

a) $\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

b) $v(t=3) = 3(3)^2 -18(3) +15=-12$

c) $t =1s, t=5s$

d) $[0,1) \cup (5,\infty)$

e) $D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

f) $a(t) = \frac{dv}{dt}= 6t -18$

g) The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

Step-by-step explanation:

For this case we have the following function given:

$f(t) = s = t^3 -9t^2 +15 t$

Part a: Find the velocity at time t.

For this case we just need to take the derivate of the position function respect to t like this:

$\frac{ds}{dt}= v(t) = 3t^2 -18t +15$

Part b: What is the velocity after 3 s?

For this case we just need to replace t=3 s into the velocity equation and we got:

$v(t=3) = 3(3)^2 -18(3) +15=-12$

Part c: When is the particle at rest?

The particle would be at rest when the velocity would be 0 so we need to solve the following equation:

$3t^2 -18 t +15 =0$

We can divide both sides of the equation by 3 and we got:

$t^2 -6t +5=0$

And if we factorize we need to find two numbers that added gives -6 and multiplied 5, so we got:

$(t-5)*(t-1) =0$

And for this case we got $t =1s, t=5s$

Part d: When is the particle moving in the positive direction? (Enter your answer in interval notation.)

For this case the particle is moving in the positive direction when the velocity is higher than 0:

$t^2 -6t +5 >0$

$(t-5) *(t-1)>0$

So then the intervals positive are $[0,1) \cup (5,\infty)$

Part e: Find the total distance traveled during the first 6 s.

We can calculate the total distance with the following integral:

$D= \int_{0}^1 3t^2 -18t +15 dt + |\int_{1}^5 3t^2 -18t +15 dt| +\int_{5}^6 3t^2 -18t +15 dt= t^3 -9t^2 +15 t \Big|_0^1 + t^3 -9t^2 +15 t \Big|_1^5 + t^3 -9t^2 +15 t \Big|_5^6$

And if we replace we got:

$D = [1 -9 +15] +[(5^3 -9* (5^2)+ 15*5)-(1-9+15)]+ [(6^3 -9(6)^2 +15*6)-(5^3 -9(5)^2 +15*5)] =7+ |32|+7 =46$

And we take the absolute value on the middle integral because the distance can't be negative.

Part f: Find the acceleration at time t.

For this case we ust need to take the derivate of the velocity respect to the time like this:

$a(t) = \frac{dv}{dt}= 6t -18$

Part g and h

The particle is speeding up $(1,3) \cup (5,\infty)$

And would be slowing down from $[0,1) \cup (3,5)$

5 0

2 years ago

The table shows the American League batting champions for the years 1999–2003. The sum of these five batting averages is 1.754.

fgiga [73]

The answer is A. .357 + .372 + .350 + .349 + x = 1.754; .326

x1 - average of 1999 Nomar Garciaparra Boston
x2 - average of 2000 Nomar Garciaparra Boston
x3 - average of 2001 Ichiro Suzuki Seattle
x4 - average of Manny Ramirez Boston
x - average of 2003 Bill Mueller Boston

The sum of these five batting averages is 1.754:
x1 + x2 + x3 + x4 + x = 1.754

From the table:
x1 = 0.357
x2 = 0.372
x3 = 0.350
x4 = 0.349
x = ?

0.357 + 0.372 + 0.350 + 0.349 + x = 1.754

1.428 + x = 1.754
x = 1.754 - 1.428
x = 0.326

5 0

3 years ago

What’s the x intercept and y intercept of 6x-5y=20

oksano4ka [1.4K]

Answer:

y-intercept equals 4 and the x-intercept is the fraction 10/3

7 0

2 years ago

the 2nd to 6th digits of the number 'b'. The number 'a': a.=.8381277236048455262835343009097967569917320013149981038679556930339

sattari [20]

Hhhhhhhhhmmmmmmmmmmmm

3 0

2 years ago