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4 years ago

Write a possible equation for a polynomial whose graph has the following horizonal intercepts. Show your work.

Mathematics

1 answer:

posledela4 years ago

6 0

Answer:

A) x - 4

B) x² - 4x

C) x³ - 2x² - 8x

D) 2x³ + 3x² - 32x + 15

Step-by-step explanation:

x-intercepts: x = 4

Factors:

(x - 4)

x-intercepts: x = 4, 0

Factor:

(x - 0)(x - 4)

x² - 4x

x-intercepts: x = 4, 0, -2

Factor:

(x - 0)(x - 4)(x + 2)

(x² - 4x)(x + 2)

x³ - 4x² + 2x² - 8x

x³ - 2x² - 8x

x-intercepts: x = 3, ½, -5

Factor:

(x - 3)(2x - 1)(x + 5)

(2x² - 6x - x + 3)(x + 5)

(2x² - 7x + 3)(x + 5)

2x³ - 7x² + 3x + 10x² - 35x + 15

2x³ + 3x² - 32x + 15

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Pls help with this for brainliest answer

TiliK225 [7]

Answer:

the answer is 12pie

Step-by-step explanation:

6mm is radius

circumference is ×2

so its 12

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3 years ago

Read 2 more answers

How many fourths equal three halves?

schepotkina [342]

2 fourths is a half so multiply 2 by 3 to get your answer that is 6

5 0

3 years ago

Pentagon A is rotated 180° about its center and then translated to create pentagon B.

liubo4ka [24]

The value of x after the simultaneous rotation and translation of pentagon A to create pentagon B is 110°

<h3>What is the translation of plane figures?</h3>

A plane figure translation is an isometry that translates each position of the figure at a distance and in a direction indicated by the vector.

It is also regarded as a sort of transformation that involves sliding each point in a plane figure at an identical distance in a similar direction.

From the attached image, the value of x after the simultaneous rotation and translation of pentagon A to create pentagon B is 110°.

Learn more about the translation of plane figures here:

brainly.com/question/2467920

4 0

2 years ago

What is the solution for this system of equations y=2/5x+4 y=2x+12

m_a_m_a [10]

2x/5 + 4 = 2x + 12
-8 = 8x/5
-8 = x

y = 2(-8) + 12
y = -16 + 12
y = -4

(-8,-4)

5 0

3 years ago

A particle moves along line segments from the origin to the points (2, 0, 0), (2, 4, 1), (0, 4, 1), and back to the origin under

Shkiper50 [21]

The work is equal to the line integral of $\vec F$ over each line segment.

Parameterize the paths

from (0, 0, 0) to (2, 0, 0) by $\vec r_1(t)=t\,\vec\imath$ with $0\le t\le2$ ,
from (2, 0, 0) to (2, 4, 1) by $\vec r_2(t)=2\,\vec\imath+4t\,\vec\jmath+t\,\vec k$ with $0\le t\le1$ ,
from (2, 4, 1) to (0, 4, 1) by $\vec r_3(t)=(2-t)\,\vec\imath+4\,\vec\jmath+\vec k$ with $0\le t\le2$ , and
from (0, 4, 1) to (0, 0, 0) by $\vec r_4(t)=(4-4t)\,\vec\jmath+(1-t)\,\vec k$ with $0\le t\le1$

The work done by $\vec F$ over each segment (call them $C_1,\ldots,C_4$ ) is

$\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r_1=\int_0^2\vec0\cdot\vec\imath\,\mathrm dt=0$

$\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec r_2=\int_0^1(t^2\,\vec\imath+24t\,\vec\jmath+32t^2\,\vec k)\cdot(4\,\vec\jmath+\vec k)\,\mathrm dt=\int_0^1(96t+32t^2)\,\mathrm dt=\frac{176}3$

$\displaystyle\int_{C_3}\vec F\cdot\mathrm d\vec r_3=\int_0^2(\vec\imath+(24-12t)\,\vec\jmath+32\,\vec k)\cdot(-\vec\imath)\,\mathrm dr=-\int_0^2\mathrm dt=-2$

$\displaystyle\int_{C_4}\vec F\cdot\mathrm d\vec r_4=\int_0^1((1-t)^2\,\vec\imath+2(4-4t)^2\,\vec k)\cdot(-4\,\vec\jmath-\vec k)\,\mathrm dt=-2\int_0^1(4-4t)^2\,\mathrm dt=-\frac{32}3$

Then the total work done by $\vec F$ over the particle's path is 46.

8 0

4 years ago