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larisa86 [58]
3 years ago
10

Write the equation in standard form of a circle with center (-3,6), tangent to the y-axis.

Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

(x+3)^2+(y-6)^2=9

Step-by-step explanation:

The equation of a circle in standard form is written as

(x-a)^2+(y-b)^2=r^2

where

(a, b) are the coordinates of the center

r is the radius of the circle

In this problem, the circle is centered at (-3,6), so we have

a = -3

b = 6

Also, we know that the circle is tangent to the y-axis, and the x-coordinate of its center is 3: this means that the horizontal distance between the point of the circle tangent to the y-axis and the centre of the circle is 3, therefore the radius of the circle is 3.

So, the equation of this circle is:

(x-(-3))^2+(y-6)^2=3^2

(x+3)^2+(y-6)^2=9

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The polynomial p(x)=3x^3-20x^2+37x-20 has a known factor of (x-4). Rewrite p(x) as a product of linear factors. p(x) =
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Answer:

p(x) = (3x^{2} - 8x + 5)(x - 4)

Step-by-step explanation:

p(x) has degree 3

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So p(x) can be rewritten as a polynomial of degree 2 multiplying (x-4)

So

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To find a, b and c, we have to divide p(x) by x - 4.

So

Finding a:

Dividing the first term of p(x) by x - 4.

\frac{3x^{3}}{x - 4} = 3x^{2}

So a = 3.

Now multiplying 3x² by, x - 4, we have:

3x^{2}(x - 4) = 3x^{3} - 12x^{2}

Subtracting p(x) from this:

3x^{3} - 20x^{2} + 37x - 20 - (3x^{3} - 12x^{2}) = 3x^{3} - 20x^{2} + 37x - 20 - 3x^{3} + 12x^{2} = -8x^{2} + 37x - 20

Finding b:

Dividing the first term, after the subtraction, by x - 4.

\frac{-8x^{2}}{x - 4} = -8x

So b = -8.

Multiplying -8x by x - 4, we have:

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Then

-8x^{2} + 37x - 20 - (-8x^{2} + 32x) = -8x^{2} + 37x - 20 + 8x^{2} - 32x = 5x - 20

Finding c:

\frac{5x}{x - 4} = 5

So c = 5.

Just to verify if the remainder is 0.

5(x - 4) = 5x - 20

5x - 20 - (5x - 20) = 0

Ok

Then:

p(x) = (ax^{2} + bx + c)(x - 4)

p(x) = (3x^{2} - 8x + 5)(x - 4)

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Step-by-step explanation:

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