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2 years ago

On a coordinate plane, a rectangle has points K prime (1, 3), L prime (2, 3), M prime (2, negative 3), N prime (1, negative 3).

Mathematics

2 answers:

cupoosta [38]2 years ago

5 0

Answer:

C and E

Step-by-step explanation: Did it on the edgenuity test. Good luck!

Sophie [7]2 years ago

3 0

Answer:

c,e

Step-by-step explanation:

right

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Irina-Kira [14]

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3 years ago

Complete the equation of the line through (-9,-9) and (-6,0)

Rudik [331]

For this case we have that by definition, the equation of the line of the slope-intersection form is given by:

$y = mx + b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis.

According to the data of the statement we have the following points:

$(x_ {1}, y_ {1}): (- 9, -9)\\(x_ {2}, y_ {2}): (- 6,0)$

We found the slope:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {0 - (- 9)} {- 6 - (- 9)} = \frac { 9} {- 6 + 9} = \frac {9} {3} = 3$

Thus, the equation is of the form:

$y = 3x + b$

We substitute one of the points and find b:

$0 = 3 (-6) + b\\0 = -18 + b\\b = 18$

Finally, the equation is:

$y = 3x + 18$

Answer:

$y = 3x + 18$

7 0

3 years ago

How many solutions would it have none infinite or one ?

Jlenok [28]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

4(5 -3y) please show steps!!

Damm [24]

Answer : 20 - 12y

Step-by-step explanation:

5 0

2 years ago

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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