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atroni [7]
2 years ago
11

Cone A is divided into Cone B and Frustum C. B A 12cm C 20cm 20cm Express the volume of Frustum C as a fraction of the volume of

Cone A. Give your answer in its simplest form.​

Mathematics
1 answer:
zaharov [31]2 years ago
7 0

The volume of the frustum is volume of the whole cone(A) minus the smaller cone(B) which is would give the volume of frustum(C) = 256cm³

<h3>Calculation of a frustum</h3>

The volume of cone A V=πr²h/3

Where radius = 20cm

The volume of cone B = V=πr²h/3

Where radius = 12cm

Therefore volume of frustum =

V=π * 20² * h/3 - π * 12² *h/3

The variables will cancel out each other

V = 20² - 12²

V = 400- 144

V = 256cm³

Therefore, the volume of the frustum(C) = 256cm³

Learn more about cone here:

brainly.com/question/1082469

#SPJ1

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Select all the rates that correspond to a unit rate of $6 per sandwich. You must select all correct responses to receive credit.
faltersainse [42]

Answer:

B, C, and E

Step-by-step explanation:

A. 42/6 = 7     Not correct

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D. 40/5 = 8     Not correct

E. 96/16 = 6     Correct

5 0
3 years ago
37​% of a certain​ country's voters think that it is too easy to vote in their country. you randomly select 12 likely voters. fi
Elenna [48]
Multiply 12 by 37%.
12 × 0.37 = 4.44

B) at least 4

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4 0
2 years ago
NO ONE EVER ANSWERS MY QUESTIONS I NEED HELP PLEASE!!! 2 problems Given: △KLM LM=12, m∠K=60°, m∠M=45° Find: Perimeter of △KLM. a
almond37 [142]

To solve this problem we will use sinus theorem.

It's general form is

a/sinα = b/sinβ = c/sinγ  

In this case it reads ML/sin∡K = KL/sin∡M = KM/sin∡L  =>

ML/sin∡K = KL/sin∡M  =>  12/sin60° = KL/sin45° => 12/(√3/2) = KL/(√2/2) =>

2*12/√3 = 2*KL/√2 => We will divide whole equation with number 2 and get

12/√3=KL/√2 => KL*√3=12√2 => KL=(12√2)/√3

When we rationalize denominator we get  KL=4√6≈ 4*2.45= 9.8

Considering that we know two angles we will calculate the third

∡K+∡M+∡L=180° => 60°+45°+∡L=180° => ∡L= 180-105=75°

Reason- The theorem of the sum of the inner angles of the triangle

ML/sin∡K=KM/sin∡L => 12/sin60°=KM/sin75°

We will calculate  sin75° = sin (30+45)= sin30 *cos45+cos30*sin45 =>

sin75°= 1/2 * √2/2 = √3/2 * √2/2= √2/4 *(√3+1)≈(1.41/4)*(1.73+1)=0.96

12/(√3/2)=KM/0.96 => 12*2/√3=KM/0.96 => 13.87=KM/0.96 =>

KM=13.87*0.96= 13.31

Perimeter of ΔKLM => P=KL+ML+KM= 12+9.8+13.31= 35.11

In the same way you can calculate perimeter of ΔMNO.

Good luck!!!


4 0
3 years ago
Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B
jeka94
\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0

M_y=35x^4+14x^5-6y^2-12xy^2
N_x=35x^4-6y^2

\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2

This suggests an integrating factor depending on x only is possible, and given by

\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}

Distributing across the ODE, we end up with

\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0

The equation is now exact, with

{M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}

Now we find the solution:

F_x=M^*
F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx
F=(7x^5y-2xy^3)e^{2x}+f(y)

F_y=N^*
(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}
f'(y)=0
\implies f(y)=C

The general solution is then

F(x,y)=(7x^5y-2xy^3)e^{2x}=C
7 0
3 years ago
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