All categories

3 years ago

Factor completely, then place the answer in the proper location on the grid. 6х^2 - 3х - 30

Mathematics

1 answer:

Tems11 [23]3 years ago

7 0

Answer:

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

Step-by-step explanation:

Given

6x² - 3x - 30 ← factor out 3 from each term

= 3(2x² - x - 10) ← factor the quadratic

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 10 = - 20 and sum = - 1

The factors are + 4 and - 5

Use these factors to split the x- term

2x² + 4x - 5x - 10 ( factor the first/second and third/fourth terms )

= 2x(x + 2) - 5(x + 2) ← factor out (x + 2) from each term

= (x + 2)(2x - 5), thus

2x² - x - 10 = (x + 2)(2x - 5) and

6x² - 3x - 30

= 3(x + 2)(2x - 5) ← in factored form

You might be interested in

Square of a standard normal: Warmup 1.0 point possible (graded, results hidden) What is the mean ????[????2] and variance ??????

LenaWriter [7]

Answer:

$E[X^2]= \frac{2!}{2^1 1!}= 1$

$Var(X^2)= 3-(1)^2 =2$

Step-by-step explanation:

For this case we can use the moment generating function for the normal model given by:

$\phi(t) = E[e^{tX}]$

And this function is very useful when the distribution analyzed have exponentials and we can write the generating moment function can be write like this:

$\phi(t) = C \int_{R} e^{tx} e^{-\frac{x^2}{2}} dx = C \int_R e^{-\frac{x^2}{2} +tx} dx = e^{\frac{t^2}{2}} C \int_R e^{-\frac{(x-t)^2}{2}}dx$

And we have that the moment generating function can be write like this:

$\phi(t) = e^{\frac{t^2}{2}$

And we can write this as an infinite series like this:

$\phi(t)= 1 +(\frac{t^2}{2})+\frac{1}{2} (\frac{t^2}{2})^2 +....+\frac{1}{k!}(\frac{t^2}{2})^k+ ...$

And since this series converges absolutely for all the possible values of tX as converges the series e^2, we can use this to write this expression:

$E[e^{tX}]= E[1+ tX +\frac{1}{2} (tX)^2 +....+\frac{1}{n!}(tX)^n +....]$

$E[e^{tX}]= 1+ E[X]t +\frac{1}{2}E[X^2]t^2 +....+\frac{1}{n1}E[X^n] t^n+...$

and we can use the property that the convergent power series can be equal only if they are equal term by term and then we have:

$\frac{1}{(2k)!} E[X^{2k}] t^{2k}=\frac{1}{k!} (\frac{t^2}{2})^k =\frac{1}{2^k k!} t^{2k}$

And then we have this:

$E[X^{2k}]=\frac{(2k)!}{2^k k!}, k=0,1,2,...$

And then we can find the $E[X^2]$

$E[X^2]= \frac{2!}{2^1 1!}= 1$

And we can find the variance like this :

$Var(X^2) = E[X^4]-[E(X^2)]^2$

And first we find:

$E[X^4]= \frac{4!}{2^2 2!}= 3$

And then the variance is given by:

$Var(X^2)= 3-(1)^2 =2$

7 0

3 years ago

If a person weighs 130 Ibs and they lose 1/10 of their weight.How much weight was lost

vampirchik [111]

Answer:

13 lbs

Step-by-step explanation:

to find 1/10 we divide by 10

This can be found by moving the decimal back by one

So 130/10=13

13 lbs were lost

8 0

3 years ago

What is the y - intercept of y-8 = 2x *

Zielflug [23.3K]

Answer:

(0,8)

Step-by-step explanation:

First, we must arrange this equation into slope-intercept form, or y=mx+b form.

Equation: y-8=2x

Add 8 to both sides: y=2x+8

Now that the equation is in slope-intercept form, it is easy to find the y-intercept, as it is just the b-value.

In this case, the b-value is 8, so the y-intercept is (0,8).

Let me know if this helps!

5 0

2 years ago

Read 2 more answers

If t and s are inversely proportional and t = 4 when s = 40, then what is t when s = 80?

Paraphin [41]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Simplify the expression. Show your work.<br><br><img src="https://tex.z-dn.net/?f=-3%284a-5b%29" id="TexFormula1" title="-3(4a-5

neonofarm [45]

Step-by-step explanation:

-3 (4a-5b)

=-3×4a -(-3×5b)

=-12a+15b

5 0

3 years ago