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3 years ago

What is the product of -2x^3+x-5 and x^3-3x-4 ?

1 answer:

6 0

To find the product of <span>-2x^3+x-5 and x^3-3x-4, we need to multiply each term in the first polynomial by the second polynomial. (So, x^3 - 3x - 4) times ....
-2x^3 = -2x^6 + 6x^4 + 8x^3
x = x^4 - 3x^2 - 4x
-5 = -5x^3 + 15x + 20
If we add all these together, we get (-2x^6 + 7x^4 + 3x^3 - 3x^2 + 11x + 20)</span>

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Write the equation of a line, in slope-intercept form, that has a slope of m=9 and passes through the point (4, 31)

Darina [25.2K]

Answer:

9x-y = 5

Step-by-step explanation:

Equation of line in slope intercept form:

y-y1 = m*(x-x1), m=9, (x1,y1)=(4,31)

y-31 = 9*(x-4)

y-31 = 9x-36

9x-y = 36-31

9x-y = 5

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2 years ago

ASAP! GIVING BRAINLIEST! Please read the question THEN answer correctly! No guessing. Show your work or give an explaination.

iogann1982 [59]

Answer:

Step-by-step explanation:

-When the function moves to the right or left to the x axis, the number has to be in "( )"

-If it moves to the right, you SUBTRACT the amount of units

-It it moves to the left, you ADD the amount of units.

So in this case, it is moving to the right 12 units so it is $(x-12)^{2}$

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3 years ago

Read 2 more answers

(m-3)/(7)=(m)/(m+8) Solve the proportion.

ExtremeBDS [4]

Answer: m=6, m=-4

Step-by-step explanation:

To solve this proportion, we have to cross multiply.

$\frac{m-3}{7} =\frac{m}{m+8}$

$(m-3)(m+8)=7m$

Now that we have cross multiplied, we actually need to FOIL the left side to expand the equation.

$m^2+8m-3m-24=7m$

Combine like terms.

$m^2+5m-24=7m$

We can move all terms to one side and then solve for m.

$m^2-2m-24=0$

We can actually factor this to:

$(m-6)(m+4)=0$

We set each factor equal to 0 to find m.

m-6=0

m=6

m+4=0

m=-4

6 0

3 years ago

Helppp please??? ill give most brainly

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Since B is the midpoint of AC, the segments AB and BC must have equal length. This means that x + 13 = 2x + 3. Solving this, we see that x = 10. By substituting 10 for x, we see that the length of AB is 23, so the length of AC is 2*23, which is 46.

5 0

3 years ago

Read 2 more answers

The random variable X has the following probability density function: fX(x) = ( xe−x , if x > 0 0, otherwise. (a) Find the mo

dusya [7]

Answer:

Follows are the solution to the given points:

Step-by-step explanation:

Given value:

$\to f_X (x) \ \ xe^{-x} \ , \ x>0$

For point a:

Moment generating function of X=?

Using formula:

$\to M(t) =E(e^{tx})= \int^{\infty}_{-\infty} \ e^{tx} f(x) \ dx$

$M(t) = \int^{\infty}_{-\infty} \ e^{tx}xe^{-x} \ dx = \int^{\infty}_{0} \ xe^{(t-1)x} \ dx$

integrating the values by parts:

$u = x \\\\dv = e^{(t-1)x}\\\\dx =dx \\\\v= \frac{e^{(t-1)x}}{t-1}\\\\M(t) =[\frac{e^{(t-1)x}}{t-1}]^{-\infty}_{0} -\int^{\infty}_{0} \frac{e^{(t-1)x}}{t-1} \ dx\\\\$

$= \frac{1}{t-1} (0) - [\frac{e^{(t-1)x}}{(t-1)^2}]^{\infty}_{0}\\\\=\frac{1}{(t-1)^2}(0-1)\\\\=\frac{1}{(t-1)^2}\\\\$

Therefore, the moment value generating by the function is $=\frac{1}{(t-1)^2}$

In point b:

$E(X^n)=?$

Using formula: $E(X^n)= M^{n}_{X}(0)$

form point (a):

$\to M_{X}(t)=\frac{1}{(t-1)^2}$

Differentiating the value with respect of t

$M'_{X}(t)=\frac{-2}{(t-1)^3}$

when $t=0$

$M'_{X}(0)=\frac{-2}{(0-1)^3}= \frac{-2}{(-1)^3}= \frac{-2}{-1}=2\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M''_{X}(t)=\frac{(-2)(-3)}{(t-1)^4}\\\\M''_{X}(0)=\frac{(-2)(-3)}{(0-1)^4}= \frac{6}{(-1)^4}= \frac{6}{1}=6\\\\M'''_{X}(t)=\frac{(-2)(-3)(-4)}{(t-1)^5}\\\\M''_{X}(0)=\frac{(-2)(-3)(-4)}{(0-1)^5}= \frac{24}{(-1)^5}= \frac{24}{-1}=-24\\\\\therefore \\\\M^{K}_{X} (t)=\frac{(-2)(-3)(-4).....(k+1)}{(t-1)^{k+2}}\\\\$

$M^{K}_{X} (0)=\frac{(-2)(-3)(-4).....(k+1)}{(-1)^{k+2}}\\\\\therefore\\\\E(X^n) = \frac{(-2)(-3)(-4).....(n+1)}{(-1)^{n+2}}\\\\$

7 0

2 years ago