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3 years ago

What are the coordinates of the vertex for f(x) = 3x2 + 6x + 3?

Mathematics

2 answers:

nikklg [1K]3 years ago

6 0

If im not wrong it would be f(x)= 6x+9
So up nine on y-axis and then go up 6 over to the right 1

Art [367]3 years ago

5 0

The vertex is (-1,0). you can find this after putting the equation in standard form by completing the square.

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Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I

Hatshy [7]

a. $f_{X,Y}$ is a joint density function if its integral over the given support is 1:

$\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1$

so the answer is yes.

b. We should first find the density of the marginal distribution, $f_Y(y)$ :

$f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy$

$f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Then

$P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}$

or about 0.2019.

For the other probability, we can use the joint PDF directly:

$P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}$

which is about 0.7326.

c. We already know the PDF for $Y$ , so we just integrate:

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5$

5 0

3 years ago

What is the vector sum of 7, 5 and 13, −5?

FrozenT [24]

The answer is (20,0). We can determine the sum of two vector points by adding x1 to x2 and y1 to y2. This is like adding integers, the only difference is the answer is in vector form. In here, we have (x,y). x = 7+13 and y = 5+(-5). that is why we have the (20,0) as the answer.

6 0

3 years ago

Read 2 more answers

Which functions have a vertex with a x-value of O? Select three options.

hram777 [196]

Answer:

Below.

Step-by-step explanation:

f(x) = |x| This is shaped like a V with the vertex at (0,0).

f(x) = |x| + 3 this is also shaped live a V with the vertex at (0, 3).

(This is |x| translated up 3 units).

f(x) = |x| - 6 similar to the above with the vertex at (0, -6).

(This is |x| translated down 6 units).

7 0

4 years ago

Need help with this math problem

Andre45 [30]

Hey There!

First you would need to factor out the two.

-4x^3/2 = -2x^3

2x^2/2= x^2

12x/2= 6x

6/2= 3

Now we have.

-2x^3+x^2+6x+3 / x+1

We would then factor with polynomial division, and then we should have.

-2(2x^2 - 3x - 3)(x+1) / x+1

We would then cancel out the x+1

Therefore, the answer would be $\boxed{-2(2x^2-3x-3)}$

Hope this helped!

8 0

4 years ago

Find m∠A. Put the answer below.<br> m∠A =

ladessa [460]

Answer:

30 degrees

Step-by-step explanation:

tanA=4√3/12

A=30

4 0

2 years ago