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3 years ago

the volume of a cone is 36π cubic units. What is the volume of a cylinder that has the same base area and the same height?

1 answer:

6 0

V_cone = 1/3 pi * r^2 * h
V_cylinder = pi*r^2*h

If you multiply the cone's volume by 3 then you get the cylinder's volume. They become the same formula
3 V_cone = pi r^2 h
Since the cone's volume is 36pi They cylinder's volume is 3*36pi = 108 pi

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Helppppppppppp plssss

Reptile [31]

Answer:

It is the 3rd choice. It is 4 times as big as the smaller cylinder

Step-by-step explanation:

V = πr2h

r - radius

h - height

π - pi

Small cyclinder

= π(3)2(10)

enter into a calucaltor and you get 282.74

Big cyclinder

= π(6)2(10)

enter into a calucaltor and you get 1130.97

Divide 1130 by 282 and you get about 4

3 0

3 years ago

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

seraphim [82]

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

$Perimeter = a\sqrt{2} + b\sqrt{10}$

Required

$a + b$

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$(x_1,y_1) = A(0,1)$

$(x_2,y_2) = B(3,4)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}$

$AB = \sqrt{( - 3)^2 + (-3)^2}$

$AB = \sqrt{9+ 9}$

$AB = \sqrt{18}$

$AB = \sqrt{9*2}$

$AB = \sqrt{9}*\sqrt{2}$

$AB = 3\sqrt{2}$

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

$(x_1,y_1) = B (3,4)$

$(x_2,y_2) = C(4,3)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}$

$BC = \sqrt{(-1)^2 + (1)^2}$

$BC = \sqrt{1 + 1}$

$BC = \sqrt{2}$

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = C(4,3)$

$(x_2,y_2) = D (3,0)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}$

$CD = \sqrt{(1)^2 + (3)^2}$

$CD = \sqrt{1 + 9}$

$CD = \sqrt{10}$

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

$(x_1,y_1) = D (3,0)$

$(x_2,y_2) = A (0,1)$

Substitute these values in the formula above

$Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}$

$DA = \sqrt{(3)^2 + (- 1)^2}$

$DA = \sqrt{9 + 1}$

$DA = \sqrt{10}$

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

$Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}$

$Perimeter = 4\sqrt{2} + 2\sqrt{10}$

Recall that

$Perimeter = a\sqrt{2} + b\sqrt{10}$

This implies that

$a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}$

By comparison

$a\sqrt{2} = 4\sqrt{2}$

Divide both sides by $\sqrt{2}$

$a = 4$

By comparison

$b\sqrt{10} = 2\sqrt{10}$

Divide both sides by $\sqrt{10}$

$b = 2$

Hence,

a + b = 2 + 10

a + b = 12

3 0

3 years ago

6x2 +10x−1 discriminant

eimsori [14]

Answer:

$124$

Step-by-step explanation:

$6 {x}^{2} +10x−1$

The discriminant of a quadratic is the expression inside the radical of the quadratic formula.

${b}^{2} −4(ac)$

Substitute in the values of a, b, and c.

${10}^{2} −4(6 \times (−1))$

$100−4(6(−1)$

$100−4(−6)$

$100 + 24$

$124$

<h3>Hope it is helpful...</h3>

8 0

3 years ago

Graph the linear equation. Find three

Karo-lina-s [1.5K]

-3y = -2x -6

Y= 2/3x + 2

When x= 3
Y= 2/3(3) + 2
Y= 2+2 = 4

(3,4)

When x= 6
Y=2/3(6)+2
Y= 4+2 = 6

(6,6)

When x= 12
Y= 2/3(12) + 2
Y= 8+2=10

(12,10)

4 0

3 years ago

Read 2 more answers

HELPPP ASAPPP DUE SOONN

Andru [333]

The friction force exerted on the 2400kg car is 9408 Newtons

<h3>What force will be exerted on the 2400kg car?</h3>

First, we know that the friction force and mass are represented by a proportional relation, this means that we can write:

F = k*M

Where F is the force, M is the mass and k is the constant of proportionality.

We know that for a 1600kg car, a force of 6272N is exerted, replacing that we get:

6272N = k*1600kg

Solving for k we get:

k = (6272N)/(1600 kg) = 3.92 N/kg

Then the proportional relationship is:

F = (3.92 N/kg)*M

So if M = 2400kg, we have:

F = (3.92 N/kg)*2400kg = 9408 N

So the friction force exerted on the 2400kg car is 9408 Newtons

If you want to learn more about proportional realtions:

brainly.com/question/12242745

#SPJ1

8 0

2 years ago