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Eddi Din [679]
3 years ago
15

Write the equation of a circle with center (6,4) that passes through the coordinate (2,1). In your final answer, include all of

your calculations.
Mathematics
2 answers:
Elodia [21]3 years ago
5 0
All points along the circle with be the distance of the radius from the center...so the radius can be found using the Pythagorean Theorem..

r^2=(4-1)^2+(6-2)^2

r^2=9+16

r^2=25

r=5

The equation of the circle can be expressed as:

r^2=(x-h)^2+(y-k)^2 where (h,k) correspond to the center of the circle, (2,1) in this case.

(x-2)^2+(y-1)^2=25

if you wanted it in a more standard form...

(y-1)^2=25-(x-2)^2

(y-1)^2=25-x^2+4x-4

(y-1)^2=-x^2+4x+21

y-1=(-x^2+4x+21)^(1/2)

y=1(+/-)(-x^2+4x+21)^(1/2)
Ainat [17]3 years ago
3 0

Answer:

(X - 6)^2 - (y - 4)^2 = 25

Step-by-step explanation:

Same as what shown except the center is (6,4 not (2,1)

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 If a^ x = b then:  

x = log_{a}b  

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8 0
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What are the x-intercepts of a parabola with a vertex of (-1,-16) and a y-intercept of (0,-15)?
Svetlanka [38]
The vertex-form equation is
  y = a(x+1)² -16
Putting in the y-intercept values, we have
  -15 = a(0+1)² -16
  1 = a . . . . . . . . . . . add 16

Then the x-intercepts can be found where y=0.
  0 = (x+1)² -16
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Do the lengths 30,40,45 form a right triangle
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5 0
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Read 2 more answers
What is 8mn⁵-2m⁶+5m²n⁴-m³n³+n⁶-4m⁶+9m²n⁴-mn⁵-4m³n³ in standar form?
vovangra [49]

The expression in standard form is n⁶ + 7mn⁵ + 14m²n⁴ - 5m³n³ - 6m⁶

<h3>How to express in standard form?</h3>

The expression is given as:

8mn⁵-2m⁶+5m²n⁴-m³n³+n⁶-4m⁶+9m²n⁴-mn⁵-4m³n³

Collect the like terms

8mn⁵ -mn⁵ - 2m⁶ -4m⁶ + 5m²n⁴ +9m²n⁴+n⁶ - 4m³n³ - m³n³

Evaluate the like terms

7mn⁵ - 6m⁶ + 14m²n⁴+n⁶ - 5m³n³

Rewrite in standard form

n⁶ + 7mn⁵ + 14m²n⁴ - 5m³n³ - 6m⁶

Hence, the expression in standard form is n⁶ + 7mn⁵ + 14m²n⁴ - 5m³n³ - 6m⁶

Read more about expressions at:

brainly.com/question/723406

#SPJ1

7 0
2 years ago
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