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Mnenie [13.5K]
3 years ago
6

How can a table represent an algebraic relationship between two variables?

Mathematics
1 answer:
frez [133]3 years ago
5 0
Here I have a video that can answer your question 

<span>https://youtu.be/iVs4dQsKBGE</span>
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Drew puechsed 3book each diffrent price for 24 doller the cost of each book was a multiple of 4. What coul be the prices of the
blsea [12.9K]
B*4+c*4+a*4=24
a=1
b=2
c=3
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5002190 in expanded firm
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5000000 000000 0000 2000 100 90 0 i hope this helps
6 0
3 years ago
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One pound of ground beef is $3.09. How much is 4.5 pounds? Round to the nearest hundredth, if necessary.
olga55 [171]

Since 1 pound is $3.09, 4.5 pounds will be:

4.5*3.09=13.905

It says to round to nearest hundredth, so your answer is $13.91.

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3 years ago
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An architect is drawing up plans for a garden that has four equal sides, each Two-fifths yard long. What is the distance around
Wittaler [7]
Answer : 1 3/5 yds
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3 0
2 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
3 years ago
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