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ratelena [41]
3 years ago
9

The variables a, b, and c represent polynomials where a = x 1, b = x2 2x − 1, and c = 2x. what is ab c in simplest form?

Mathematics
2 answers:
inna [77]3 years ago
7 0
If you would like to write a * b + c in simplest form, you can do this using the following steps:

a = x + 1
b = x^2 + 2x - 1
c = 2x
a * b + c = (x + 1) * (x^2 + 2x - 1) + 2x = x^3 + 2x^2 - x + x^2 + 2x - 1 + 2x = x^3 + 3x^2 + 3x - 1

The correct result would be x^3 + 3x^2 + 3x - 1.




QveST [7]3 years ago
5 0

easyer way its c...........

...........

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baherus [9]
I got -64/3 for this problem.
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3 years ago
The area of the triangle above will equal one half of a rectangle that is 5 units long and units wide. (Input only whole numbers
il63 [147K]

Answer:

<h2>Width of the rectangle is 2</h2>

Step-by-step explanation:

A rectangle can is made up of two right angled triangle as shown.

Since area of a rectangle = base * height

Area of one of the triangle = \frac{1}{2}\  * \ area\ of\ the\ rectangle

The length and width of the triangle will be the same as that of the rectangle i.e  5units long and 2units wide

Area of one of the triangle = \frac{1}{2}  * 5 * 2

Area of one of the triangle = 5units²

The area of the triangle above will equal one half of a rectangle that is 5 units long and 2 units wide.

5 0
3 years ago
Please help me!!!
nikitadnepr [17]

Answer:

B

Step-by-step explanation:

The answer is option B

When drawn option B looks like the image

it crosses at the points (-1,-3), (0,-2),(2,0)

rearranging the equations and drawing a table graph shows how to draw the lines

5 0
4 years ago
For what value of k is there one solution to the given quadratic equation (k+1)x²+4kx+2=0.
andriy [413]

Answer:

If k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

Step-by-step explanation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

The signal of \bigtriangleup determines how many real roots an equation has:

\bigtriangleup > 0: Two real and different solutions

\bigtriangleup = 0: One real solution

\bigtriangleup < 0: No real solutions

In this problem, we have the following second order polynomial:

(k+1)x^{2} + 4kx + 2 = 0.

This means that a = k+1; b = 4k; c = 2

It has one solution if

\bigtriangleup = 0

b^{2} - 4ac = 0

16k^{2} -8(k+1) = 0

16k^{2} - 8k - 8 = 0

We can simplify by 8

2k^{2} - k - 1 = 0

The solution is:

k = 1 or k = -\frac{1}{2}

So, if k = 1 or k = -\frac{1}{2}, there is only one solution to the given quadratic equation.

4 0
3 years ago
Log32- log 32-log 4. simplify​
Rufina [12.5K]

Answer:

huh

Step-by-step explanation:

...

..

.

.

.

.............

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3 years ago
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