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NeX [460]
3 years ago
11

Write a polynomial functionſ of least degree that has rational coefficients, a leading coefficient of 1, and the zeros 1,3, and

Mathematics
1 answer:
otez555 [7]3 years ago
8 0

Answer:

f(x) = x^{4}  -  8x^{3}  + 24x^{2}  - 32x  + 15

Step-by-step explanation:

First of all, if 2 - i is a zero, then so is 2 + i

All you need to do now is multiply (x - 1)(x - 3)(x - 2 + i)(x - 2 - i)

I will do part of that multiplication

(x - 2 + i)(x - 2 - i) = x^{2} - 2x - xi - 2x + 4 + 2i + xi - 2i - i^{2}

                           = x^{2} - 4x + 5

So, now (x - 1)(x - 3)(x^{2} - 4x + 5) =  x^{4}  -  8x^{3}  + 24x^{2}  - 32x  + 15

I will let you finish that multiplication.  I did some and you can do some.  OK?

                                     

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Montano1993 [528]
<h3><u>Answer</u><u>:</u><u>-</u></h3>

192 cm²

<h3><u>Step</u><u> </u><u>by</u><u> step</u><u> explanation</u><u> </u><u>:</u><u>-</u></h3>

Let us take the height be x , then its side = x + 4. Now half of base will be 12 cm .

<u>According</u><u> to Pythagoras Theorem :- </u>

=> base² + perpendicular ² = hypontenuse ²

=> 12² + x² = (x+4)²

=> 144 + x² = x² + 16 + 8x

=> 8x = 144-16

=> 8x = 128

=> x = 128/8

=> x = 16 cm .

Hence the height of ∆ is 16 cm .So the area will be half the product of base and altitude.

= 1/2 * 16 cm * 24cm .

= 192 cm²

<h3><u>★</u><u> </u><u>Hence</u><u> </u><u>the</u><u> </u><u>area</u><u> </u><u>of</u><u> </u><u>the </u><u>tria</u><u>ngle</u><u> is</u><u> </u><u>1</u><u>9</u><u>2</u><u> </u><u>cm²</u><u> </u><u>.</u></h3>
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3 years ago
What is the perpendicular slope to 6/-3
IRISSAK [1]

Answer:

Perpendicular slope is \frac{3}{6} or \frac{1}{2}.

Step-by-step explanation:

Given:

Let the perpendicular slope be x.

The given slope is \frac{6}{-3}

For perpendicular slopes, the product of the slopes is equal to -1.

Therefore, x\times \frac{6}{-3}=-1\\-\frac{6}{3}x=-1\\6x=3\\x=\frac{3}{6}=\frac{1}{2}

Therefore, the perpendicular slope is \frac{3}{6} or \frac{1}{2}.

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4 years ago
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125 divided by 5 is 25. hope this helped but im not to sure what friendly parts are sorry
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3 years ago
Write the equation of a Line that is perpendicular to y=5/2x+3 and passes
koban [17]

Answer:

Step-by-step explanation:

Slope of perpendicular lines = -1

y = 5/2x + 3

m_{1}=\frac{5}{2}\\\\m_{1}*m_{2}=-1\\\\

        m_{2}= -1 ÷ m_{1}

               = -1*\frac{2}{5}=\frac{-2}{5}

(-3 , -5)

Equation of the required line: y - y₁ = m(x -x₁)

                                                y - [5] = \frac{-2}{5}(x - [-3])\\

                                               y+5=\frac{-2}{5}x + 3*\frac{-2}{5}\\\\y+5=\frac{-2}{5}x-\frac{6}{5}\\\\   y =\frac{-2}{5}x-\frac{6}{5}-5\\\\   y = \frac{-2}{5}x-\frac{6}{5}-\frac{5*5}{1*5}\\\\ y = \frac{-2}{5}x-\frac{6}{5}-\frac{25}{5}\\\\\\ y=\frac{-2}{5}x-\frac{31}{5}

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4 years ago
(-3,600,000,000)x(0.00000824)
Nadusha1986 [10]
-29664 is the correct answer. This mostly deals with the zeros and crossing them off
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