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sashaice [31]
3 years ago
15

What is 8 to the power of 4?

Mathematics
2 answers:
Dafna11 [192]3 years ago
6 0

Answer:

4096

Step-by-step explanation:

Akimi4 [234]3 years ago
3 0
(8 • 8)(8 • 8)
= 64 • 64
=4096
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Describe the location of the point having the following coordinates. negative abscissa, zero ordinate between Quadrant II and Qu
Marianna [84]

Answer:

The location of the point is between Quadrant II and Quadrant III

Step-by-step explanation:

we know that

The abscissa refers to the x-axis  and ordinate refers to the y-axis

so

in this problem we have

the coordinates of the point are (-x,0)

see the attached figure to better understand the problem

The location of the point is between Quadrant II and Quadrant III

6 0
3 years ago
The quotient of 3 times a number and 7 is 15. Find the number<br> pls help me
Ronch [10]

Answer:

it would be 3 times 35 divded by 7 and that answer would give you 15 as the answer so 35 is your missing number

Step-by-step explanation:

4 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
Graph the equation below by plotting the y-intercept and a second point on the line. (Zoom in if blurry)
stira [4]

Answer:

The coordinates I chose were (0,2) and (4,5)

Step-by-step explanation:

Didn't you already do this question?

6 0
3 years ago
Read 2 more answers
7x(-3) x (-2)^2<br> help please!
Anvisha [2.4K]
    [7×(-3)]   ×   (-2)²
=     (-21)    ×    4
=             -84
5 0
3 years ago
Read 2 more answers
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