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BartSMP [9]
3 years ago
9

Can I get some help with the first one. 6p^3-12p^2+9p

Mathematics
1 answer:
Gelneren [198K]3 years ago
8 0
Hi there!

The answer is:
6p {}^{3}  - 12 {p}^{2}  + 9p = 3p(2 {p}^{2}  - 4p + 3)

We can find this answer by writing the factors.
6 {p}^{3}  = 3 \times 2 \times p \times  p\times p
- 12 {p}^{2}  = 3 \times  - 4 \times p \times p
9p = 3 \times 3 \times p

Now we can see that the terms that are in common are 3 and p. Therefore, the GCF is 3p.

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Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)
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Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

x^{2} +y^{2}+4x-1=0

Step-by-step explanation:

<u>Explanation:</u>-

<u>Step 1:</u>-

The equation of the circle having center and radius is

(x-h)^2+(y-k)^2=r^2

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

      (\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2}  )

    (-2,0)

therefore center (h,k) = (-2,0)

<u>Step 2:-</u>

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}

r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}

r=\sqrt{5}

<u>Step 3</u>:-

center (h,k) = (-2,0) and

radius r=\sqrt{5}

The standard form of circle equation

(x-h)^2+(y-k)^2=r^2

(x-(-2))^2+(y-0)^2=\sqrt{5} ^2

on simplification is

x^{2} +y^{2}+4 x-1=0

5 0
3 years ago
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