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tino4ka555 [31]
2 years ago
13

You buy 9 pizzas. Your total cost is $57.89 with an additional charge of $12 for tip. Write an equation that represents the scen

ario. Use c for the unknown cost of each pizza.​
Mathematics
1 answer:
ki77a [65]2 years ago
7 0

Answer:

<h3>57.89 - 12 ÷ 9= c </h3><h3>im pretty sure !!!!</h3><h3>step-by-step explanation:</h3>

I HOPE THIS HELPS YOU OUT WITH YOUR QUESTION! PLZ LET ME KNOW IF IT DID! IT MAKES ME FEEL GOOD THAT SOMEONE APPRECIATES IT! ANYWAY, IF YOU HAVE A FEW MINUTES PLZ GOT TO MY PROFILE AND THANK MY ANSWERS (THE MORE THE MERRIER)

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Subtract 6 from me. Then multiply by 2. If you subtract 40 and then divide by 4, you get 8. What number am I?
aliina [53]

Answer:

42

Step-by-step explanation:

Let the number be "x"

We translate the word equation to algebraic equation.

First,

subtract 6 from me, so we have:

x - 6

Now,

Multiply by 2, so we have:

2(x-6)

Now,

Subtract 40, then divide by 4:

\frac{2(x-6)-40}{4}

Now, you get "8", so equate this to 8, we get:

\frac{2(x-6)-40}{4}=8

We now solve using algebra:

\frac{2(x-6)-40}{4}=8\\2(x-6)-40=4*8\\2(x-6)-40=32\\2x-12-40=32\\2x-52=32\\2x=32+52\\2x=84\\x=42

The number is 42

7 0
3 years ago
How do i find the average speed for his question: "A bicycle racer rides from a starting marker at 10m/s. She then rides back al
padilas [110]

To find the average speed for the whole race, you just have to add the speed from a starting marker and the speed from the turnaround marker going back and divide it by 2 since there are only two speeds involved.

10 m/s + 16m/s = 26 m/s / 2 = 13 m/s

Therefore, the average speed for the whole race is 13 m/s.

7 0
3 years ago
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What is the graph of this equation ​
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Answer:

Look at attached image

Step-by-step explanation:

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3 years ago
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Ainat [17]
The answer would be the first one or 3y2, this is because it goes into each term evenly enough to simplify it.
6 0
3 years ago
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Solve the proportion.
Marysya12 [62]

Answer:

y = 5

Step-by-step explanation:

cross-multiply:

6y = 30

y = 5

3 0
3 years ago
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