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2 years ago

WILL MARK BRAINLIEST!! PLZ HELP!!! 20 POINTS Create a graph of the polynomial function g(x) = (x + 2)(x − 1)(x − 2)

Mathematics

1 answer:

Gennadij [26K]2 years ago

6 0

Answer:

We have: g(x)=(x+2)(x-1)(x-2)

multiply the 3 terms:

(x+2)(x-1)(x-2)
(x²-4)(x-1)
x^3 -x²-4x+4

This is a polynimial function with 2 vertices

The roots are approximatively:

0.8 and -1.5

their images are:

1.9 and 8.8

plot these two vertices and draw some other points

here is a drawing:

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Rewrite 25 and 125 as exponents that have the same base. Write your answers #^# and

dangina [55]

Answer:

25 = 5^2
125 = 5^3

Step-by-step explanation:

Both 25 and 125 are powers of 5. As such, they can be written as the powers they are:

25 = 5·5 = 5^2

125 = 25·5 = 5·5·5 = 5^3

3 0

2 years ago

Wright the expression 6a - 2 (a-1) in simplest form

vovikov84 [41]

6a-2(a-1)
6a-2a-2
4a-2
I think it is 4a-2

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3 years ago

Yo i need help please i’ll give a brainliest to whoever gets the correct answer

kipiarov [429]

Answer:

Student B because an irrational number cannot be of the form p/q

Step-by-step explanation:

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3 years ago

Please help me with geometry! <3<br> In rhombus ABCD, m<=53*, What is M (*=degrees)

julia-pushkina [17]

In a rhombus opposite angles are equal and all 4 add up to 360
assuming one angle is 53, the opposite angle will also be 53 and then adjacent angles will be (360 - (53 x 2))/2 = 254/2 = 127*(deg)
I think the answer you're looking for is 127

3 0

2 years ago

Find the general solution to 3y′′+12y=0. Give your answer as y=... . In your answer, use c1 and c2 to denote arbitrary constants

lozanna [386]

Answer:

$y(x)=c_1e^{2ix}+c_2e^{-2ix}$

Step-by-step explanation:

You have the following differential equation:

$3y''+12y=0$ (1)

In order to find the solution to the equation, you can use the method of the characteristic polynomial.

The characteristic polynomial of the given differential equation is:

$3m^2+12=0\\\\m^2=-\frac{12}{3}=-4\\\\m_{1,2}=\pm2\sqrt{-1}=\pm2i$

The solution of the differential equation is:

$y(x)=c_1e^{m_1x}+c_2e^{m_2x}$ (2)

where m1 and m2 are the roots of the characteristic polynomial.

You replace the values obtained for m1 and m2 in the equation (2). Then, the solution to the differential equation is:

$y(x)=c_1e^{2ix}+c_2e^{-2ix}$

4 0

3 years ago