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Tom [10]
4 years ago
9

What is 130000 milliliters in liters?

Mathematics
1 answer:
e-lub [12.9K]4 years ago
3 0
I think its 130 Liters
bcz idk if I do this right
1000=1 Liter
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Given that ∠XQR = 180° and ∠LQM = 180°, which equation could be used to solve problems involving the relationships between ∠MQR
evablogger [386]
The answer is B, when you add ∠MQR and ∠LQR you will get <span>180°.</span>
5 0
3 years ago
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Find two numbers, if their sum is 93 and their difference is 9.
ivolga24 [154]

Answer:

51 and 42

Step-by-step explanation:

Use

x + y = 93

x - y = 9

Add those two equations together and you get 2x = 102

*y is not in the equation anymore because y + -y = 0*

Divide both sides by 2

x = 51

plug x into one of the equations; let's use x + y = 93

Then you get, 51 + y = 93

Subtract both sides by 51 and you get y = 42

Plug both in the other equation to make sure the numbers work.

51 + 42 = 93; true

51 - 42 = 9; true

4 0
3 years ago
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???<br><br>a) y = 3x – 1<br><br>b) y = -2x + 1<br><br>c) g(x) = 3x + 6<br><br>​
Goryan [66]
A+ B equals C times G squared divided by four gives you the dividend of 9×8
3 0
4 years ago
Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
inessss [21]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

x + 0 y - 3 z = 7 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Subtract 1/2 × (equation 1) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y/2 - 2 z = 3/2 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

Subtract 1/3 × (equation 2) from equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by 2:

{x+0 y+0 z = 10 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0
3 years ago
Simplify (y x y^1/3)^3/2
jok3333 [9.3K]

Answer:

y^2

Step-by-step explanation:

(y x y^1/3)^3/2 would be clearer if written as

(y*y^(1/3))^(3/2).

Because y^a*y^b = y^(a + b), the quantity inside the first set of parentheses becomes

y^(4/3)

and if we apply the power (3/2) to this result, we get:

                                 4       3

(y^(4/3))^(3/2) = y^{ ----- * ----- ) = y^2

                                 3       2

3 0
3 years ago
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