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viktelen [127]
3 years ago
9

Write a polynomial in standard form with zeros 1, 2, and −5.

Mathematics
1 answer:
Sonbull [250]3 years ago
6 0

Answer:

Step-by-step explanation:

First thing you have to do is write those zeros into factors.  If

x = 1, then its factor is (x - 1).  If

x = 2, then its factor is (x - 2).  If

x = -5, then its factor is (x + 5).

Now we FOIL them all together, starting with the first 2:

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

Combine like terms to get

x^2-3x+2

Now FOIL that with the last factor to get

x^3-3x^2+2x+5x^2-15x+10

Combine like terms to get

x^3+2x^2-13x+10=0, which is choice C

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Step-by-step explanation:

The area of a disk of revolution at any x about the x- axis is πy² where y=2x. If we integrate this area on the given range of values of x from x=0 to x=1 , we will get the volume of revolution about the x-axis, which here equals,

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Now we have to calculate the volume of revolution about the y-axis. For that we have to first see by drawing the diagram that the area of the CD like disk centered about the y-axis for any y, as we rotate the triangular area given in the question would be pi - pi*x². if we integrate this area over the range of value of y that is from y=0 to y=2 , we will obtain the volume of revolution about the y-axis, which is given by,

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\huge {625}^{ \frac{3}{8} }  \times  {5}^{ \frac{1}{2} }  \div 25 \\  \\ =  \huge {( {5}^{4})}^{ \frac{3}{8} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\  =   \huge{5}^{ \cancel4 \times  \frac{3}{ \cancel8 \:  \:  \red{ \bold 2}} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\   \huge=  {5}^{ \frac{3}{2} }  \times  {5}^{ \frac{1}{2} }  \div  {5}^{2}  \\  \\ \huge=  {5}^{ \frac{3}{2}  + \frac{1}{2}}  \div  {5}^{2}  \\ \\ \huge=  {5}^{ \frac{4}{2}  }  \div  {5}^{2}  \\ \\ \huge=  {5}^{ 2  }  \div  {5}^{2}  \\  \\   \huge=  {5}^{ 2  - 2 }  \\  \\   \huge=  {5}^{ 0 }  \\  \\   \huge=  1

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3 years ago
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