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KATRIN_1 [288]
3 years ago
9

Item 9 A company that offers tubing trips down a river rents tubes for a person to use and “cooler” tubes to carry food and wate

r. A group spends $270 to rent a total of 15 tubes. Write a system of linear equations that represents this situation. Use xx to represent the number of one-person tubes rented and yy to represent the number of cooler tubes rented. How many of each type of tube does the group rent?
Mathematics
1 answer:
s2008m [1.1K]3 years ago
7 0
You would need to know how much it cost to rent each type of tube in order to solve this problem.

One of the equations would be xx+yy=15
The number of single tubes plus the number of cooler tubes is 15.

If you know how much each tube cost, simply put the cost in front of the xx/yy and set it equal to 270$. For example, if the single tubes cost 15$ and the cooler tubes cost 20$, then the equation would be 15xx+20yy=270
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Linear Relationships Study Guide 1.) Select all the equations for which (-6, -1) is a solution. Show your work to prove that eac
TiliK225 [7]

we have point (-6, - 1)

Now we will put these points in each equation,

y = 4x +23

put x = -6 and y = -1

-1 = 4 (-6) +23

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-1 = -1

LHS = RHS, so this equation has (-6 , -1) as solution.

y = 6x

put x = -6 and y = -1

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LHS is not equal RHS, so (-6 , -1) is not a solution for that equation,

y = 3x - 5

put x = -6 and y = -1

-1 = 3 (-6) - 5

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y= 1/6 x

put x = -6 and y = -1

-1 = -6/6

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LHS = RHS, so (-6 , -1) is a solution for that equation,

6 0
1 year ago
(1/27)^x-6=27<br><br>Could I please have a step by step as well? Much appreciated! ​
Andrei [34K]

Answer:

x = 7

Step-by-step explanation:

1) Use Division Distributive Property: (x/y)^a = x^a/y^a.

\frac{1}{ {27}^{x - 8} }  = 27

2) Multiply both sides by 27^x - 8.

1 =  {27} \times 27^{1 + x  - 8}

3) Use the product rule: x^a x^b = x^a+b.

1 =  {27}^{1 + x - 8}

4) Simplify 1 + x - 8 to x - 7.

1  =  {27}^{x - 7}

5) Use Definition of Common Logarithm: b^a = x if and only if log<u>b</u><u> </u>(x) = a.

log_{27}1 = x - 7

6) Use Change of Base Rule.

\frac{ log_1  }{  log{27} }  = x - 7

7) Use rule of 1: log 1 = 0.

\frac{0}{ log_{27}}  = x - 7

8) Simplify 0/log_27 to 0.

0 = x - 7

9) Add 7 to both sides.

7 = x

10) Switch sides.

x = 7

<u>Therefor</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>x</u><u> </u><u>=</u><u> </u><u>7</u><u>.</u>

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