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3 years ago

What is the product (3x-5)(2x^2-7x+1)

Mathematics

1 answer:

timurjin [86]3 years ago

3 0

Answer:

= 6x^3-33x^2+45x-6

Step-by-step explanation:

= (3x+-6)(2x^2+-7x+1)

= (3x)(2x^2)+(3x)(-7x)+(3x)(1)+(-6)(2x^2)+(-6)(-7x)+(-6)(1)

= 6x^3-21x^2+3x-12x^2+42x-6

= 6x^3-33x^2+45x-6

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-18m^3n^6/-6mn^3 simplify

IRINA_888 [86]

Answer:

$3m^{2}n^{3}$

Step-by-step explanation:

we have:

$\frac{-18m^{3}n^{6}}{-6mn^{3}} = 3m^{(3-1)}*n^{(6-3)}= 3m^{2}n^{3}$

7 0

3 years ago

Whats a interger greater than 10 and less than -15 ?

mestny [16]

An integer greater than 10 is 46.

An integer less than -15 is -24.

7 0

4 years ago

Graph the following pair of quadratic functions and describe any similarities/differences observed in the graphs.

Degger [83]

The difference between the two graphs is that the first one is increasing which means that it’s going positive infinity. While the second graph is decreasing which means that the errors are going negative infinity.

F(x)= 8x^2+2

6 0

3 years ago

Please help :c the first word is find

xeze [42]

Answer:

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

Step-by-step explanation:

The complete statement is:

Find $\alpha_{2}$ and $r_{2}$ such that $\sin \theta - \sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos (\theta - \alpha_{2})$ .

We proceed to use the following trigonometric identity:

$\cos (\theta - \alpha_{2}) = \cos \theta \cdot \cos \alpha_{2} +\sin \theta \cdot \sin \alpha_{2}$ (1)

$\sin \theta -\sqrt{3}\cdot \cos \theta = r_{2}\cdot \cos \theta \cdot \cos \alpha_{2}+r_{2}\cdot \sin \theta \cdot \sin \alpha_{2}$

By direct comparison we derive these expressions:

$r_{2}\cdot \sin \alpha_{2} = 1$ (2)

$r_{2}\cdot \cos \alpha_{2} = -\sqrt{3}$ (3)

By dividing (2) by (3), we have the following formula:

$\tan \alpha_{2} = -\frac{1}{\sqrt{3}}$

$\tan \alpha_{2} = -\frac{\sqrt{3}}{3}$

The tangent function is negative at second and fourth quadrants. That is:

$\alpha_{2} = \tan^{-1} \left(-\frac{\sqrt{3}}{3} \right)$

There are at least two solutions:

$\alpha_{2,1} = 150^{\circ}$ , $\alpha_{2,2} = 330^{\circ}$

And the value of $r_{2}$ :

$r_{2}^{2}\cdot \sin^{2}\alpha_{2} + r_{2}^{2}\cdot \cos^{2}\alpha_{2} = 4$

$r_{2}^{2} = 4$

$r_{2} = 2$

The values of $r_{2}$ and $\alpha_{2}$ are 2 and 150º.

5 0

4 years ago

Helpp me plzzz im being timed

drek231 [11]

Above question
Slope = change in y / change in x
(2 different random point on the line)
Let’s point out 2 points: (0,3) and (6,0)
So slope = (3 - 0) / (0 - 6) = 3/(-6) = -1/2

Y-intercept is when x = 0, the line cross the y-axis at y = ?
So y = 3
Below question is farmiliar:
Slope = (30-50) / (0-6) = -20/(-6) = 10/3
Y-intercept = 30

3 0

3 years ago

Read 2 more answers