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faust18 [17]
3 years ago
15

Find three consecutive integers with a sum of 267

Mathematics
2 answers:
Tanya [424]3 years ago
6 0
88 + 89 + 90 = 267 and they are consecutive integers as they are right next to each other.

the easiest way to solve this problem is to divide 267/3 to get 89, and then use this information to gather your three integers that should be of about the same value.
Bad White [126]3 years ago
6 0
267 divided 3 equals 89 so the numbers average to 89.

So 88,89,90 are the only ones that average to 89 and add up to 267

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Can anyone help solve this for me ASAP. Thanks!
Anuta_ua [19.1K]

Answer:

x = 20

y = 26

Step-by-step explanation:

40 and 2x are vertical angles, so they're equal to each other.

40 = 2x

Divide both sides by 2

x = 20

We see that 40 and 5y + 10 are a linear pair, which means they add up to 180.

5y + 10 + 40 = 180

5y + 50 = 180

5y = 130

y = 26

8 0
3 years ago
What is the point of comparing prices?
kherson [118]

Answer:

To know which is better to pay.

Step-by-step explanation:

Good Luck

6 0
3 years ago
Read 2 more answers
Mrs. Gilus is in charge of the budget for the school. Last year, we began the year with &708.88 in our bank account, but by
liberstina [14]
It sounds like they had $708.88 at the beginning, however by the end of the year they OWED $347.99. The first number is in the positive while the other is a negative. Add the last number to the first number for the amount that they have spent which would be $1,056.87.
8 0
3 years ago
for the school play adult tickets cost 4$ and children tickets cost 2$ natalie is working at the ticket counter and just sold 20
lilavasa [31]
Let us formulate the independent equation that represents the problem. We let x be the cost for adult tickets and y be the cost for children tickets. All of the sales should equal to $20. Since each adult costs $4 and each child costs $2, the equation should be

4x + 2y = 20

There are two unknown but only one independent equation. We cannot solve an exact solution for this. One way to solve this is to state all the possibilities. Let's start by assigning values of x. The least value of x possible is 0. This is when no adults but only children bought the tickets.

When x=0,
4(0) + 2y = 20
y = 10

When x=1,
4(1) + 2y = 20
y = 8

When x=2,
4(2) + 2y = 20
y = 6

When x=3,
4(3) + 2y= 20
y = 4

When x = 4,
4(4) + 2y = 20
y = 2

When x = 5,
4(5) + 2y = 20
y = 0

When x = 6,
4(6) + 2y = 20
y = -2

A negative value for y is impossible. Therefore, the list of possible combination ends at x =5. To summarize, the combinations of adults and children tickets sold is tabulated below:

   Number of adult tickets             Number of children tickets
                  0                                                   10
                  1                                                    8
                  2                                                    6
                  3                                                    4
                  4                                                    2
                  5                                                    0




6 0
4 years ago
ASAP <br>What is the domain?
WARRIOR [948]

real numbers hope it helps

7 0
3 years ago
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