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Norma-Jean [14]
2 years ago
12

The junior senator can work his way through

Mathematics
1 answer:
Gre4nikov [31]2 years ago
7 0

Answer:

It is supposed to be 25

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2 ^( 3 - 9) - 11 how do I solve this
Setler [38]
2^(3-9)-11 =

1 step: subtract the 3 and 9 to get 6

(3-9)=6

2 step: multiply 2 to its self 6 times and
you should get 64

2*2=4*2=8*2=16*2=32*2=64

3 step: subtract 11 by 64 to get 53

64-11=53

hope I help if so hit the brainliest thanks
5 0
3 years ago
Jessie earns $12 an hour.How much will he earn in 12 hours
Stels [109]

Answer:

$144

Step-by-step explanation:

since Jessie earns $12 an hour, if he works 12 hours, multiply the amount per hour by the amount of hours worked.

12 x 12 = 144

Therefore, Jessie will earn $144 in 12 hours

8 0
2 years ago
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What is the surface area of a cylinder with base radius 4 and height 5
crimeas [40]
20 because radius times height will give you the surface area of the shape
6 0
3 years ago
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Please help me I’m terrible at math
irga5000 [103]

Answer:

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Step-by-step explanation:

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8 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
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