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3 years ago

If x+y=12and x-y=8, what is the value of 2x-xy?

Mathematics

1 answer:

slamgirl [31]3 years ago

8 0

Answer:

2x - xy = 0

Step-by-step explanation:

We need to solve the equations simultaneously to find the values of x and y:

x + y = 12

x - y = 8 - we need to add the equations together

2x = 20

x = 10

Then we substitute x into one of the equations to find y:

10 + y = 12

12 - 10 = y

y = 2

Now we substitute x and y into the final expression:

2(10) - 2(10)

20 - 20 = 0

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6a+4=61/3+4<br> I don’t know what I’m doing wrong because I keep getting 10 1/3

Vlad1618 [11]

Answer:

a = 3 7/18 or 3.388889

Step-by-step explanation:

First, you want to get the a by itself, so you would subtract 4 from the left side of the equals sign.

6a + 4 = 61 / 3 + 4

-4 -4

6a = 61 / 3

By doing that, they cancel each other out, so you are left with

6a = 61 / 3

Then you want to get a by itself, so you divide 6 from both sides.

6a / 6 = 61 / 3 / 6

Remember that 6 can also be written as 6/1

Remember that dividing a fraction is multiplying the reciprocal.

6a / 6 = 61 / 3 / 6 / 1

When dividing fractions remember to Skip the first fraction, Flip the second fraction, and Multiply the two fractions together.

61 / 3 x 1 / 6

skip multiply flip

Multiply across the top, and across the bottom.

61(1) and 3(6)

Then you get

61 / 18

So a = 61 / 18

If you want to convert that to a mixed number, it would be 3 7 / 18

If you want that in a decimal, it would be 3.388889

4 0

2 years ago

Find the perimeter of the following triangle.

pashok25 [27]

Hello!

To find the perimeter of the triangle, we need to find the length of all the sides using the <u>distance formula</u>.

The distance formula is: $d =\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

First, we can find the distance between the points (-3, -1) and (2, -1). The point (-3, -1) can be assigned to $(x_{1},y_{1})$ , and (2, -1) is assigned to $(x_{2},y_{2})$ . Then, substitute the values into the formula.

$d =\sqrt{(2 - (-3))^{2}+ (-1 - (-1))^{2}}$

$d =\sqrt{5^{2}+0^{2}}$

$d =\sqrt{25} = 5$

The distance between the points (-3, -1) and (2, -1) is 5 units.

Secondly, we need to find the distance between the points (2, 3) and (2, -1). Assign those points to $(x_{1},y_{1})$ and $(x_{2},y_{2})$ , then substitute it into the formula.

$d =\sqrt{(2 - 2)^{2}+ (-1 - 3)^{2}}$

$d =\sqrt{0^{2}+(-4)^{2}}$

$d =\sqrt{16} = 4$

The distance between the two points (2, 3) and (2, -1) is 4 units.

Finally, we use the distance formula again to find the distance between the points (-3, -1) and (2, 3). Remember the assign the ordered pairs to $(x_{1},y_{1})$ and $(x_{2},y_{2})$ and substitute!

$d =\sqrt{(2 -(-3))^{2}+ (3 - (-1))^{2}}$