All categories

2 years ago

Easy question:

Mathematics

1 answer:

ivolga24 [154]2 years ago

6 0

Answer:

midpoint = (-2 , 6)

Step-by-step explanation:

The midpoint can be found with the formula {(x1 + x2 )/2, (y1 + y2.) P1( x1 , y1 ), and P2 (x2 , y2 ) are the coordinates of two points, and the midpoint is a point lying equidistant and between these two points.

You might be interested in

Find the zero of the function f(x) = -8x + 4.<br> NEED HELP FAST DONT HAVE MUCH TIME

Paladinen [302]

Answer:

1/2

Step-by-step explanation:

5 0

2 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

2 years ago

A model length of 12 cm. represents an actual length of 102 ft.the model has a width of 48cm. What is the width of the actual ob

Elodia [21]

the answer would be 408

just set it up like this

12 48

---- = ----

102 ?

12*4=48 so 102*4 = 408

8 0

3 years ago

Find the center and radius of the with the equation (x + 5) ^ 2 + (y - 2) ^ 2 = 25

iVinArrow [24]

Answer: Center: (5,0)

Radius:5

Step-by-step explanation: Happy to help :)

8 0

3 years ago

Factor completely. 12x4+18x3+45x2 Enter your answer in the box.

lukranit [14]

QUESTION 1

The given expression is

$12x^4+18x^3+45x^2$

The greatest common factor is $3x^2$ .

We factor to obtain;

$3x^2(4x^2+6x+15)$

QUESTION 2

The given quadratic equation is

$x^2+6x-72=0$

We split the middle term to obtain

$x^2+12x-6x-72=0$

Factor by grouping;

$x(x+12)-6(x+12)=0$

$(x+12)(x-6)=0$

Use zero product property;

$(x+12)=0\:\:or\:\:(x-6)=0$

$x=-12\:\:or\:\:x=6$

QUESTION 3

The given system of equation is

$3x+4y=-8$

$7x-3y=6$

If we multiply $3x+4y=-8$ by 3, we obtain;

$9x+12y=-24$

If we multiply $7x-3y=6$ by 4 we obtain;

$28x-12y=24$

Adding the last two equations will give us;

$37x=0$

The y-variable is eliminated.

Answer:Multiply 3x+4y=−8

by 3. Multiply 7x−3y=6 by 4. Add the resulting equations together.

6 0

3 years ago