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makvit [3.9K]
3 years ago
10

In May, Jim’s lunch account has a balance of $58.19. If lunch costs $2.74 per day, how many days will Jim be able to buy lunch b

efore his account runs out of money?
Mathematics
1 answer:
kykrilka [37]3 years ago
6 0

Answer:

<u>21 days.</u>

Step-by-step explanation:

To solve this, divide the cost of lunch from the initial amount in his account. Therefore:

58.19 = 2.74x ('x' being the number of days)

Divide both sides by 2.74:

\frac{58.19}{2.74} = \frac{2.74x}{2.74}

x ≈21.24, or 21 days.

***Since you use whole numbers when counting days, round down to determine the amount of days he can buy lunch. Thus, he can buy lunch for <u>21 days.</u>

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the second longest bridge in the world, the tianjin grand bridge, is 373,000 feet long. Its length is 167,700 feet less than the
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4 0
3 years ago
-8z+(4.5)+3.5z+7y-1.5
bogdanovich [222]

Answer:

  • \boxed{\sf{7y+3-4.5z}}

Step-by-step explanation:

In order to combine like terms, you have to isolate x and y from one side of the equation.

\sf{-8z+\left(4.5\right)+3.5z+7y-1.5}

<u>First, thing you do is remove parentheses.</u>

\Longrightarrow: \sf{-8z+4.5+3.5z+7y-1.5}

<u>Solve.</u>

<u>Then, you combine like terms.</u>

\Longrightarrow:\sf{-8z+3.5z+7y+4.5-1.5}

<u>Add/subtract the numbers from left to right.</u>

-8z+3.5z=-4.5z

<u>Rewrite the problem down.</u>

\Longrightarrow: \sf{-4.5z+7y+4.5-1.5}

<u>Solve.</u>

4.5-1.5=3

\Longrightarrow: \boxed{\sf{7y+3-4.5z}}

  • <u>Therefore, the correct answer is 7y+3-4.5z.</u>

I hope this helps, let me know if you have any questions.

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2 years ago
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