All categories

3 years ago

Solve this plz, why does this have to have 20 characters

Mathematics

1 answer:

miskamm [114]3 years ago

8 0

2y^2 + 4y + 6 + y^2 + y

2y^2 + y^2 + 4y + y

3y^2 + 5y

You might be interested in

I really need help. I've checked everywhere and it says cannot be factored. I will give brainliest to whoever can help me and gi

jolli1 [7]

Answer:

B) -5, 5, -7, 7

Step-by-step explanation:

3x² - bx + 2

factor:

(3x - 2 )(x - 1) = 3x² - (5)x + 2 so 5 does work

(3x - 1 )(x - 2) = 3x² - (7)x + 2 so 7 does work

(3x + 2 )(x + 1) = 3x² + 5x +2 = 3x² - (-5)x + 2 so -5 does work

(3x + 1 )(x + 2) = 3x² + 7x + 2 = 3x² - (-7)x + 2 so -7 does work

4 0

3 years ago

What is the volume of this cone?

Katyanochek1 [597]

There is no image but I would likely be glad to help! :)

4 0

3 years ago

Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago

A plant grows at a rate of 7 ft every two weeks

vlabodo [156]

Answer:

0 is 0,1 is 7,2 is 14 and 3 is 20 (c. is that it has not grown yet and (d. is 7 per week

3 0

3 years ago

Read 2 more answers

What is the formula to get the volume of a cylinder

bezimeni [28]

Answer:

V = π × r² × h

Step-by-step explanation:

All you need to do is multiply the base times the height and since the base is a circle whose formula is π × r², we get π × r² × h.

5 0

3 years ago

Read 2 more answers