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Anit [1.1K]
3 years ago
8

Find the volume of the cone

Mathematics
1 answer:
4vir4ik [10]3 years ago
6 0

Answer:

12\pi units^3

Step-by-step explanation:

V=1/3Bh

=1/3 Area of the base times height

Area of base: \pir^2=4\pi

height=9

4\pi times 9=36\pi

1/3 of 36\pi =12\pi

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Evaluate 6x + 11 when x = 7
nlexa [21]

Answer:

53

Step-by-step explanation:

To evaluate 6x + 11, we must substitute the value of x.

Since we know that x = 7, it is easier to evaluate the expression;

6x + 11

6(7) + 11

Since 6 is outside the parenthesis, we must multiply everything inside the parenthesis by 6;

6(7) + 11

42 + 11

When you add the two numbers you get:

53

4 0
3 years ago
A cross-country racer travels 20 kilometers before she realizes that she has to cover 75 kilometers in order to qualify for the
Katarina [22]
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3 years ago
The Atlas V Rocket could go 1 full mile in just
Jet001 [13]

Answer: It traveled 36,000 miles per hour.

Step-by-step explanation: 1/10 of a second x 36000 = 1 hour.

5 0
3 years ago
Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
3 years ago
Solve for x: −10x+5= -6x-35
Makovka662 [10]
X= 10 I did it on my calculator so don’t worry :)
5 0
3 years ago
Read 2 more answers
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