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murzikaleks [220]
2 years ago
12

The radius of the Ferris wheel is shown. Which is the circumference of the Ferris wheel, rounded to the nearest foot?

Mathematics
2 answers:
Yuri [45]2 years ago
8 0

Answer:

About 785.4

Step-by-step explanation:

C = 2pir

R = 125

2*pi*125 ≈ 785.4

C ≈ 785.4

:) Have a good day :)

mixas84 [53]2 years ago
7 0

Answer:

The circumference is around 785 feet

Step-by-step explanation:

The circumference of a circle is pi x radius x 2 or pi x diameter.

If the Ferris wheel's radius is 125, then this is what you do:

pi x 125 x 2

= pi x 250

= 3.14 x 250

= 785

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7 0
3 years ago
Subtract -2k3 + k2 - 9 from 5k3 - 3k + 7.
tankabanditka [31]
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Evaluate the integral. 3 2 t3i t t − 2 j t sin(πt)k dt
sveta [45]

∫(t = 2 to 3) t^3 dt

= (1/4)t^4 {for t = 2 to 3}

= 65/4.

----

∫(t = 2 to 3) t √(t - 2) dt

= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2

= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du

= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}

= 26/15.

----

For the k-entry, use integration by parts with

u = t, dv = sin(πt) dt

du = 1 dt, v = (-1/π) cos(πt).


So, ∫(t = 2 to 3) t sin(πt) dt

= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt

= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]

= 5/π + 0

= 5/π.

Therefore,

∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.

3 0
3 years ago
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