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ELEN [110]
3 years ago
5

A student got 78 marks out of a possible 120. Express the students mark as a percentage

Mathematics
2 answers:
statuscvo [17]3 years ago
8 0

Answer:

65%

Step-by-step explanation:

78/120=13/20

13/20=65/100

65/100=65%

sasho [114]3 years ago
5 0
The answer is 65%


(78/120)*100

=65%
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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>
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3 years ago
Expand, and simplify the following expressions a. (2x+y)(x+y) + (2x-y)(x+y)​
tankabanditka [31]

Step-by-step explanation:

2x²+2xy+yx+y²+2x²+2xy-yx+y²

2x²+2x²+2xy+2xy+yx-yx+y²+y²

2x²+2x²+2xy+2xy

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2 years ago
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Shenny earned $25 after babysitting for 3 hours ​
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Answer:

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Step-by-step explanation:

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Match solutions and differential equations. Note: Each equation may have more than one solution. Select all that apply. (a) 7y''
Nonamiya [84]

Answer:

a-e^x,e^{-x}

b-x^{-2}

c-x^3

Step-by-step explanation:

a.7 y''-7 y =0

Auxillary equation

D^2-1=0

(D-1)(D+1)=0

D=1,-1

Then , the solution of given differential equation

y=e^x,y=e^{-x}

2.7x^2y''+14xy'-14 y=0

Y=y=x^3

y=3x^2

y''=6x

Substitute in the given differential equation

42x^3+42x^3-14x^3\neq 0

Hence, x^3 is not a solution of given differential equation

e^x,e^{-x} are also not a solution of given differential equation.

y=x^{-2}

y'=-2x^{-3}

y''=6x^{-4}

Substitute the values in the differential equation

7x^2(6x^{-4})+14x(-2x^{-3})-14 x^{-2}

=42x^{-2}-28x^{-2}-14x^{-2}=0

Hence, x^{-2} is a solution of given differential equation.

c.7x^2y''-42y=0

y=x^3

y'=3x^2

y''=6x

Substitute the values in the differential equation

42x^3-42x^3=0

Hence, x^3 is a solution of given differential equation.

a-e^x,e^{-x}

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c-x^3

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Answer:

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Step-by-step explanation:

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