MATHEMATICS HIGH SCHOOL

Which polynomial function has a leading coefficient of 2, root -4 with multiplicity 3, and root 10 with multiplicity 1?O f(x) = 2(x-4)(x-4)(x - 4)(x + 10)

O fx) = 2(x + 4)(« + 4)(« + 4)(-10)

O f) =3K – 4)(«-4)0x + 10)

O f()= 3(x + 4)(x + 4)(K- 10)

Answers

Answer 1
Answer:

Answer:

f(x) = 2(x+4)(x+4)(x+4)(x-10)

Step-by-step explanation:

Here, we want to write a polynomial function with leading coefficient of 2, root - 4 with multiplicity 3 ( meaning it appears 3 times);

and root 10 with multiplicity 1

If x = -4

Then in linear form, we have x + 4

if x = 10, in linear form, we have x -10

So the polynomial is;

f(x) = 2(x+4)(x+4)(x+4)(x-10)

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Replace the? with one of following symbols (<, >, =, or) 2+3? 4+1
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What is the value of -4 + 91 + 72?

Provide an Example of how to draw transformed figures that are translated, reflected, and rotated.

Answers

Step-by-step explanation:

Translations: When a figure is slid to another place on the coordinate grid (for example if i asked you to move the figure 2 units down and 4 units to the left)

Reflected: When a figure is reflected across the y-axis or the x-axis and both figures are congruent to each other.

Rotated: When a figure is rotated (for example if i asked you to rotate the figure 90 degrees)

Find the equation of the circle whose center and radius are given.center ( -2, -5), radius = 1

Answers

Answer:  The required equation of the circle is x^2+y^2+4x+10y+28=0.

Step-by-step explanation:  We are given to find the equation of the circle with center (-2, -5) and radius of length 1 unit.

We know that

the standard equation of a circle with center (h, k) and radius of length r units is given by

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

For the given circle, we have

center, (h, k) = (-2, -5)   and  radius, r = 1 units.

Therefore, from equation (i), we get

(x-(-2))^2+(y-(-5))^2=1^2\n\n\Rightarrow (x+2)^2+(y+5)^2=1\n\n\Rightarrow x^2+4x+4+y^2+10x+25=1\n\n\Rightarrow x^2+y^2+4x+10y+29-1=0\n\n\Rightarrow x^2+y^2+4x+10y+28=0.

Thus, the required equation of the circle is x^2+y^2+4x+10y+28=0.
as i said before put the center instead of (h,k) in the general formula and put r=1 
so (x-(-2))^2 + (y-(-5)^2 = 1 
(x+2)^2 + (y+5)^2 = 1

What is the first 5 common multiples of 15 and 25

Answers

You can first find the missing factors of each number to find the lcm
So 15 = 5*3 and 25 = 5*5
In order to make them equal, Multiply 15 by 5 and 25 by 3 so that both of them are 5*3*5 which equals 75.
After that you can just multiply each by 2, 3, 4, and 5 because all the other common multiples are multiples of 75. You get
150, 225, 300, 375, as well as 75
So:
75
150
225
300
375
15 = 15,30,45,60,75,90,105,120,135,150,165,180,195,210,225,240, 255, 270,285,300,315,330,345,360,375
25 = 25,50,75,100,125,150,175,200,225,250,275,300,325,350,375

The first 5 common multiples of 15 and 25 are 75, 150, 225, 300, and 375.

0.55 rounded to the nearest tenth

Answers

0.60 because if the number of the right it’s 5 or more you change the number of the left and all the numbers to the right change to #0

f(x) = x³ - 3x² + 3x - 1f(x + 1) What is the value?

A) x³+1
B) x³-1
C) x³
D) x²
E) x²+1

Answers

f(x+1)=(x+1)^3-3(x+1)^2+3(x+1)-1\n f(x+1)=x^3+3x^2+3x+1-3(x^2+2x+1)+3x+3-1\n f(x+1)=x^3+3x^2+6x+3-3x^2-6x-3\n f(x+1)=x^3 \Rightarrow C
f(x)=(x-1)³ = x³ -3x² +3x -1 
f(x+1) = (x+1-1)³ = x³

Solve each quadratic equation by factoring.0=8a² -10a+3

Answers

8a² - 10a + 3 = 0
x = -10 +/- √(-10)² - 4(8)(3)
                      2(8)
x = -10 +/- √100 - 96
                 16
x = -10 +/- √4
            16
x = -10 +/- 2
           16
x = -10 + 2    x = -10 - 2
           16                 16
x = -8            x = -12
      16                   16
x = -1/2        x = -3/4
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