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Afina-wow [57]
3 years ago
12

Solve the system of equations below. х– Зу =- 4 Зх – 9y =- 12

Mathematics
1 answer:
OleMash [197]3 years ago
5 0
For the first equation it is 0. For the second equation it is 0 again.. am I right?
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A community center rents their hall for special events. They charge a fixed fee of $200 plus an hourly fee of $22.50. Lin has $3
Karo-lina-s [1.5K]

Answer:

No, 6 hours would cost $335, 35 over her limit of $300.

Step-by-step explanation:

she has to pay $200 plus 22.50 for every hour she rents. So, if she rents for 6 hours, she has to pay $200 + 6(22.50) or $200 + 135=335. She only has $300 so she can’t afford the event.

3 0
2 years ago
Devide 60 in the ratio of 2:3​
nevsk [136]

Answer:

Let the amounts be 2x and 3x.

Then 2x+3x=60

So, 5x=60

So, x=60/5=12

So, amounts are

2x=2*12=24

And 3x=3*12=36

Hope this helps you!

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Number 2 I need help on how to solve it
Hunter-Best [27]

Let's simplify the following expression:

\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}

We get,

\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{  }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}

4 0
1 year ago
Use order of operation
Inessa [10]
(6+23)=29
(32-25)=7
(29)x(7)+(7x7)
203+49= 252
the answer is 252
6 0
3 years ago
What is 6 1/2 - 2 2/5 - 2 1/2 =
VikaD [51]

Answer:

1 3/5

Step-by-step explanation:

Change to mixed fractions, that'll give:

13/2 - 12/5 - 5/2

Using lcm:

16/10

Change back to mixed fractions:

1 6/10

Lowest term:

1 3/5

4 0
3 years ago
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