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3 years ago

How to find radius of base of cone

Mathematics

1 answer:

algol133 years ago

3 0

Answer:

The radius of a cone is the radius of its circular base. You can find a radius through its volume and height. Multiply the volume by 3. For example, the volume is 20.

I hope it's helpful!

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bonufazy [111]

Answer:

see explanation

Step-by-step explanation:

Under a rotation about the origin of 90°

a point (x, y ) → (- y, x ), thus

A(2, 2 ) → A'(- 2, 2 )

B(2, 4 ) → B'(- 4, 2 )

C(4, 6 ) → C'(- 6, 4 )

D(6, 4 ) → D'(- 4, 6 )

E(6, 2 ) → E'(- 2, 6 )

6 0

3 years ago

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BartSMP [9]

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7 0

3 years ago

Use the graph to answer the question.

dmitriy555 [2]

Answer:

Step-by-step explanation:

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3 0

3 years ago

Test the series for convergence or divergence (using ratio test)

Triss [41]

Answer:

$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

3 0

3 years ago

If B is the midpoint of AC, and AC=8x-20. Find BC AB=3x-1 Show all work

exis [7]

First we know that:
AB+BC=AC
now , since B is the midpoint of AC, this means that AB=BC
therefore,
AB+BC=AC can also be written as
AB+AB=AC
3x-1+3x-1=8x-20
6x-2=8x-20
-2+20=8x-6x
18=2x
x=9

therefore:
AB=3(9)-1=27-1=26
AC=8(9)-20=52
BC=AB=26

6 0

3 years ago

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