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4 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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10 + 6x + 2x - 7 = 35<br> Help me pls and ty

lutik1710 [3]

Answer:

Explanation:

10 + 6x + 2x - 7 = 35

Add Similar Elements: 6x + 2x = 8x

10 + 8x - 7 = 35

Group Like Terms

8x + 10 - 7 = 35

Subtract: 10 - 7 = 3

8x + 3 = 35

Subtract 3 From Both Sides

8x + 3 - 3 = 35 - 3

Simplify

8x = 32

Divide Both Sides By 8

8x / 8 = 32 / 8

6 0

3 years ago

Read 2 more answers

Q no 13 and 14 answer

GalinKa [24]

Answer:

<h2>Question 13 : x = 62°</h2>

<h2>Question 14 : x = 2</h2>

Step-by-step explanation:

Question 13

The diagram in question 13 is a triangle.

The sum of interior angles of a triangle is equal to 180°

To solve for x ;

$x + 69 + 49 = 180 \\ x + 118 = 180 \\ x = 180 - 118 \\ x = 62$

Question 14

(: represents ratio or /)

x/9 = 18/81

Cross Multiply

81x = 18×9

81x = 162

Divide both sides of the equation by 81

81x/81 = 162/81

x = 2

7 0

3 years ago

Help your girl out!

Sergio [31]

The correct answer is A

6 0

3 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

$T(1)=\frac{}{2+1}=\frac{1}{3}$

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

12x^2-17xy-5y^2 factorise

melisa1 [442]

Answer:

(4x + y)(3x-5x)

Step-by-step explanation:

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3 years ago