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3 years ago

Plz help me I am having a lot of trouble

Mathematics

2 answers:

MakcuM [25]3 years ago

4 0

3/4 / 5/1= 3/20 i hope this answer helps you

Sedaia [141]3 years ago

3 0

300 pounds? I can only think in pounds for each bed. I hope that helps?

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What is the greatest common factor of the following terms?<br> 20x4y3, 26x5y3 and 12x2y3

Lena [83]

Answer:

2x²y³

<em>good luck, i hope this helps :)</em>

8 0

3 years ago

Given the set of vertices, determine whether parallelogram ABCD is a rhombus, rectangle or square. List all that apply. A(7,-4),

Sloan [31]

Given:

Vertices of a parallelogram ABCD are A(7,-4), B(-1,-4), C(-1,-12), D(7, -12).

To find:

Whether the parallelogram ABCD is a rhombus, rectangle or square.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using distance formula, we get

$AB=\sqrt{(-4-(-4))^2+(-1-7)^2}$

$AB=\sqrt{(-4+4)^2+(-8)^2}$

$AB=\sqrt{0+64}$

$AB=8$

Similarly,

$BC=\sqrt{(-1-(-1))^2+(12-(-4))^2}=8$

$CD=\sqrt{(7-(-1))^2+(-12-(-12))^2}=8$

$AD=\sqrt{(7-7)^2+(-12-(-4))^2}=8$

All sides of parallelogram are equal.

$AC=\sqrt{(-1-7)^2+(-12-(-4))^2}=8\sqrt{2}$

$BD=\sqrt{(7-(-1))^2+(-12-(-4))^2}=8\sqrt{2}$

Both diagonals are equal.

Since, all sides are equal and both diagonals are equal, therefore, the parallelogram ABCD is a square.

We know that, a square is special case of rectangles and rhombus.

So, parallelogram ABCD is a rhombus, rectangle or square. Therefore, the correct option is c.

7 0

3 years ago

Plz explain your answer.

klio [65]

Number six question is answers “Mid point”

7 0

3 years ago

What are rights a teacher has

Lorico [155]

Answer:

Teachers have the right to be free from discrimination based on race, sex, and national origin. As well as freedom of expression, academics, privacy, and religion.

Good luck! Hope this helped :)

3 0

3 years ago

Read 2 more answers

Lydia graphed ADEF at the coordinates D (-2,-1), E (-2, 2), and F (0,0). She thinks ADEF is a right triangle. Is Lydia's asserti

Setler79 [48]

The true statement is (c) No; the slopes of segment EF and segment DF are not opposite reciprocals.

<h3>Right triangles </h3>

Right triangles have a pair of perpendicular lines

Coordinates

The coordinates are given as:

D = (-2,-1)
E = (-2,2)
F = (0,0)

<h3>Slopes</h3>

Start by calculating the slopes of lines DF and EF using:

$m = \frac{y_2 -y_1}{x_2 -x_1}$

So, we have:

$m_{DF} = \frac{0 + 1}{0 +2}$

$m_{DF} = \frac{1}{2}$

Also, we have:

$m_{EF} = \frac{0 -2 }{0+2}$

$m_{EF} = \frac{-2 }{2}$

$m_{EF} = -1$

For the triangle to be a right triangle, then the calculated slopes must be opposite reciprocals.

i.e.

$m_1 = -\frac{1}{m_2}$

By comparison, the slopes of both lines are not opposite reciprocals.

Hence, the true statement is (c)