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Dmitriy789 [7]
3 years ago
5

I am between 7,000,000 and 8,000,000. all my digits are odd. all the digits in my thousands period are the same. the sum of my d

igit is 31. what number am i? give as many answers as you can. what strategies did you use to find the mystrey number
Mathematics
2 answers:
lozanna [386]3 years ago
7 0

Answer:

The number can be many numbers like: 7,777,111  or 7,555,333  or 7,111,777

A: 7,777,111

We have four 7's and three 1's

7*4=28

1*3=3

28+3=31

B: 7,555,333

We have one 7, three 5's and three 3's.

5*3=15

3*3=9

15+9+7=31

These also can be 7,333,555 or 7,111,777.

aniked [119]3 years ago
4 0
There are many possibilities so I am going to give 3 answers 777113 777131 777311
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Answer:

D. 67

Step-by-step explanation:

To help me see the region we needed to look at, I wrote the inequalities out.

Let x be number of pre-sale tickets and y be the number of at-the-door tickets as your graph suggests.

So one of the inequalities about number of where the other one is about cost.

You are given x+y is no more than 400 or x+y<=400 (the top line graphed in your picture is x+y=400).

You are given 10x+25y is at least 5000 or 10x+25y>=5000 (the bottom line graphed in your picture).

I solved both of these for y.

x+y<=400

Subtract x on both sides giving y<=-x+400 (shaded below line because of the y< part).

10x+25y>=5000

Subtract 10x on both sides:

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Divide both sides by 25:

          y>=-2/5 x+200 (shaded above the line because of y> part).

The region we should then be looking at is:

Let's look at the points (0,400), (0,200), and finally (333,67).

Cost=10x+25y

Let's plug in

Cost=10(0)+25(400)=10000

Cost=10(0)+25(200)=5000  (can we go lower than 200)

Cost=10(333)+25(67)=5005 (our y is lower 200 so far this is the winner)

Cost 10(334)+25(66)=4990 (didn't meet the 5000 dollar requirement)

67 D.

(Also if you look at the graph 66 would not be included in the shaded region; it would be too low)

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