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Nezavi [6.7K]
3 years ago
15

Tomas bought 80 tickets for rides at an amusement park. Each ride costs 5 tickets, and Tomas has been on x rides so far. Which e

xpression is equivalent to the number of tickets that Tomas has left? Select all that apply. A. 80 – 5x B. 80 + 5x C. 5(16 – x) D. –5x + 80 E. 5(16 + x)
Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

Option A, C and D are correct choices.

Step-by-step explanation:

Let x be the number of rides that Tomas has been on so far. Each ride costs 5 tickets. This means that cost of x rides will be 5x.

We are told that Tomas bought 80 tickets for rides at an amusement park.

To find the number of of tickets that Tomas has left we will subtract cost of x rides from total number of tickets that Thomas bought.

We can represent this information in an expression as: 80-5x

Now let us see which of our given choices in equivalent to our expression.

A. 80-5x

Upon looking at option A we can see that it is same as our expression, therefore, option A is the correct choice.

B. 80+5x

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option B is not a correct choice.

C. 5(16-x)

Upon distributing 5 we will get,

80-5x

Now this expression is same same as our expression, therefore, option C is the correct choice.

D. -5x+80

This expression is also same as our expression as we can rearrange the terms in this expression as: 80-5x, therefore, option D is a correct choice as well.

E. 5(16+x)

Upon distributing 5 we will get,

80+5x

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option E is not a correct choice.

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4t+3c=$81.00<br> 10t+3c=$135.00
muminat
\left \{ {{4t+3c=81.00\:(I)} \atop {10t+3c=135.00\:(II)}} \right.
<span>Simplify the first equation by (-1)
</span>\left \{ {{4t+3c=81.00\:*(-1)} \atop {10t+3c=135.00\:\:\:\:\:}} \right.
<span>Once simplified, cancel the opposing terms.
</span>\left \{ {{-4t-\diagup\!\!\!\!3c=-81.00} \atop {10t+\diagup\!\!\!\!3c=135.00}} \right.
Now find the value of "t".
\left \{ {{-4t=-81.00} \atop {10t=135.00}} \right.
-------------------------------
6t = 54.00
t =  \frac{54.00}{6}
\boxed{\boxed{t = 9.00}}\end{array}}\qquad\quad\checkmark

Now find the value of "c", replace the found value of "t" in the first equation:
4t+3c=81.00\:(I)
4*9+3c=81.00
36 + 3c = 81.00
3c = 81.00 - 36
3c = 45.00
c =  \frac{45.00}{3}
\boxed{\boxed{c = 15.00}}\end{array}}\qquad\quad\checkmark

Answer:
<span>The values ​​of "t" and "c" satisfying the linear system are successively: $ 9.00 and $ 15.00</span>
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Answer:

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