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nordsb [41]
3 years ago
9

5. Lauren has $50 to spend at the Strawberry Festival. She spends $ 10 on admission and $12 on strawberry shortcake. She wants t

o play an arcade game that costs $0.75 per game. Write an inequality to represent Lauren's situation. Find the maximum number of times, n, Lauren can play the game. Justify your answer.
Mathematics
1 answer:
olga55 [171]3 years ago
4 0

For this case we have that Lauren's initial amount of money was $ 50, if he spends $10 and $12 on admission and strawberry pie respectively we have:

$50 - ($10 + $12) = $50- $22 = $28

That is, she has $ 28 left

n: Let the variable that represents the amount of games Lauren can play.

Then, it must be fulfilled that:

0.75n \leq28\\n \leq \frac {28} {0.75}\\n \leq37.33

Thus, Lauren can play 37 times the game (at most)

ANswer:

n \leq37.33

Thus, Lauren can play 37 times the game (at most)

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3 years ago
Need help with my homework ​
Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

8 0
3 years ago
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