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3 years ago

The volume of a cone is 16 cubic inches. Its height is 12 inches. What is the radius of the cone?

Mathematics

1 answer:

Ivan3 years ago

8 0

Answer:

r ≈ 1.13 inches

Step-by-step explanation:

if i round it i guess the answer is 2

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Please can someone help me.

Vladimir79 [104]

Answer:

Step-by-step explanation:

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3 years ago

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Consider the given function and the given interval. f$x$ = 2 sin$x$ - sin$2 x$ text(, ) [0 text(, ) pi] (a) Find the avera

Finger [1]

a. The average value of $f$ on the given interval is

$\displaystyle f_{\rm ave}=\frac1{\pi-0}\int_0^\pi(2\sin x-\sin2x)\,\mathrm dx=\boxed{\frac4\pi}$

b. Solve for $c$ :

$\dfrac4\pi=2\sin c-\sin2c\implies\boxed{c\approx1.238\text{ or }c\approx2.808}$

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3 years ago

Please help, I don’t understand it so explain don’t just put the answer if u can lols

ludmilkaskok [199]

Answer

27 days

Step-by-step explanation:

father= 45 days

oldest son + father=27 days

youngest + father= 36 days

45-27=18

45-36=9

18+9=27

so it would take 27 days for both sons to complete the house

18 days=oldest son

9 days =youngest son

7 0

4 years ago

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Bicyclist mike hall rode around the world in 91 days.what fraction is the simplest form represents the part of a year in weeks h

victus00 [196]

52 weeks
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3 years ago

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Expand the given power using the Binomial Theorem. (10k – m)5

agasfer [191]

Answer:

$(10k - m)^{5}=100000k-50000k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

Step-by-step explanation:

* Lets explain how to solve the problem

- The rule of expand the binomial is:

$(a+b)^{n}=(a)^{n}+nC1(a)^{n-1}(b)+nC2(a)^{n-2}(b)^{2}+nC3(a)^{n-3}(b)^{3}+...............+(b)^{5}$

∵ The binomial is $(10k-m)^{5}$

∴ a = 10k , b = -m and n = 5

∴ $(10k-m)^{5}=(10k)^{5}+5C1(10k)^{4}(-m)+5C2(10k)^{3}(-m)^{2}+5C3(10k)^{2}(-m)^{3}+5C4(10k)^{1} (-m)^{4}+5C5(10k)^{0}(-m)^{5}$

∵ 5C1 = 5

∵ 5C2 = 10

∵ 5C3 = 10

∵ 5C4 = 5

∵ 5C5 = 1

∴ $(10k-m)^{5}=100000k^{5}+(5)(10000)k^{4}(-m)+(10)(1000)k^{3}(m^{2})+(10)(100)k^{2}(-m^{3})+5(10k)^{1} (m^{4})+(10k)^{0}(-m^{5})$

∴ $(10k-m)^{5}=100000k^{5}-50000)k^{4}m+10000k^{3}m^{2}-1000k^{2}m^{3}+50km^{4}-m^{5}$

5 0

3 years ago