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3 years ago

a family went out to dinner and the cost of the food was $80 .If the trip given was 15%, what was the total cost

Mathematics

2 answers:

frutty [35]3 years ago

7 0

12 dollars.

80/10=8
8/2=4
8+4=12

12345 [234]3 years ago

6 0

Hi! :)

So i assume you meant the tip, not trip, but that's fine.

An easy way to calculate the tip is to move the decimal on the percentage left two spaces, then multiply the decimal by the food cost.

For example, in this case you would make the 15% into .15 and multiply .15 by 80.

.15*80=12.

This means you would leave a $12 tip, and to get your total bill just add the tip to the food cost.

12+80=92

So your total would be $92

Hope this helps!

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yarga [219]

37 and 1/2 could mean:
37+ (1/2)
37 times 1/2

37+1/2=37 and 1/2 or 37.5

37 times 1/2=37/2=18 and 1/2=18.5

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4 years ago

5=4b-12 hurry and make sure you put how you got the answer

PIT_PIT [208]

Answer:

4.25

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3 years ago

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3. Two linear functions are shown below. Which function has the greater rate of change? Justify your response LOOK AT PIC! plz h

Rainbow [258]

Answer:

Function B has the greater rate of change

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7 0

3 years ago

Joan took her three children to the zoo. Admission is $7 each. The snack bar sells water for $2 per bottle and chips for $3 per

lys-0071 [83]

The expression for the total cost of their trip is 7 + 2x + 3y.

As per the question, assuming the number of bottles to be x and number of bags to be y. The equation will have one time fee of admission, product of cost of bottles and number of bottles and product of cost of bags and number of chips bags.

The one time fee and two products will be added to form the expression. Forming the equation now -

Expression = 7 + 2x + 3y

Therefore, the expression of trip when x bottles of water were bought during the visit is 7 + 2x + 3y.

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8 0

1 year ago

find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large

katrin2010 [14]

A generic odd number can be written as

$2k+1,\quad k \in \mathbb{Z}$

Since there is an odd number every two numbers, three consecutive odd numbers will be

$2k+1,\quad 2k+3,\quad 2k+5$

Now let's make up the equations: the sum of the first two is

$(2k+1)+(2k+3)$

And 27 less than 3 times the largest is

$3(2k+5)-27$

These two must be the same, so we have

$(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3$

Subtracting 4k and 3 from both sides gives

$1=2k \iff k=\dfrac{1}{2}$

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0

4 years ago