3. Translate the triangle 3 units left

Rectangle ABCD translates 4 units down and 2 units to the right to form rectangle A'B'C'D'. The vertices of rectangle ABCD are l

AlekseyPX

The length of B'C' in the rectangle A'B'C'D' = 9 units.

<u>Step-by-step explanation</u>:

step 1 :

Draw a rectangle with vertices ABCD in clockwise direction.

where, AB and DC are width of the rectangle ABCD.

AD and BC are length of the rectangle ABCD.

step 2 :

Now,

The length of the rectangle is AD = 5 units and

The width of the rectangle is AB = 3 units.

step 3 :

Draw another rectangle with vertices A'B'C'D' extended from vertices of the previous rectangle ABCD.

step 3 :

The length of the new rectangle is A'D' which is 4 units down from AD.

∴ The length of A'D' = length of AD + 4 units = 5+4 = 9 units

step 4 :

Since B'C' is also the length of the rectangle A'B'C'D', then the measure of B'C' is 9 units.

3 0

3 years ago

Can someone please help me

Oksana_A [137]

The answer is -31 over 12.

5 0

2 years ago

Read 2 more answers

Find the slope of the line passing through (-9,3), (-3,-5)

egoroff_w [7]

Answer:

-4/3

Step-by-step explanation:

Slope: (y2-y1)/(x2-x1)

(-5-3)/(-3+9) = -8/6 = -4/3

The slope is -4/3

6 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

2 years ago

The total cost of five pens and eight mechanical pencils is 12.65. The cost of each mechanical pencil is 0.75 cents. Which repre

Tanya [424]

Answer:

the cost of each pen is 1.33

Step-by-step explanation:

if you multiply 0.75 by 8 you get 6 after that you subtract 6 from 12.65 which means you have 6.65 which you divide by 5 which equals 1.33

7 0

3 years ago