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3 years ago

CAN SOMEONE PLEASE HELP ME ON THIS

Mathematics

1 answer:

PIT_PIT [208]3 years ago

6 0

I think that you just have to put the x values to the equation for example :

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HELP PLEASE!!!!!! Find x and y so the quadrilateral is a parallelogram.

adell [148]

Answer:

x is on R,Q and Q,P and y is R,S to S,P

Step-by-step explanation:

picture it on a coordinate plane

8 0

3 years ago

Simplify 5m-2(6m-5)+2 a. -7m-8 b. -7m+12 c. 17m+12 d. 17m-8

fgiga [73]

The answer is B. -7m+12.

First distribute the -2 to 6m and -5.

5m-2(6m-5)+2

5m+(-2*6m)+(-2*-5)+2

5m+-12m+10+2

After that, combine the like terms.

5m+-12m+10+2

-5m+-12m=-7m

10+2=12

The simplified expression is -7m+12.

7 0

4 years ago

What is 1/4 times 48

svetlana [45]

Simplifiy the answer it'll become 2

8 0

4 years ago

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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

Here is an expression: 3 • 2t

leva [86]

Answer:

1. 6

2. 24

Step-by-step explanation:

Substitute in the values for t and solve the expression.

3 0

3 years ago

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