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12345 [234]
3 years ago
10

How many tables are needed ~~~~~PLEASE HELP

Mathematics
1 answer:
12345 [234]3 years ago
8 0

Answer:

it would be 5 tables

Step-by-step explanation:

1:6

2:10

3:14

4:18

5:22

just count by 4

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Draw a graph for the situation.
n200080 [17]
It will be a discrete graph, where there is no dependant nor independent variables, they are not related by any means.

Hope this helps.

6 0
3 years ago
jenny went to the mall with $120 in her wallet.she spent $25.65 on a shirt, $7.98 on a pair of socks, $16.23 on a cd , and $4.58
Effectus [21]
Total money=$120
on shirt=$25.65
on socks=$7.98
on a cd=$16.23
on snack=$4.58
so total money left=total money-total money spent 
=$120-($25.65+$7.98+$16.23+$4.58)
=$120-$54.44
so total money left=$65.56
4 0
3 years ago
Read 2 more answers
Emily wants to watch her intake of fat per day.
AfilCa [17]

Answer:

One variable equation that is (4800/x) represents percentage of Emily's dinner fat intake compared to total daily allowance of x gram.

Step-by-step explanation:

lets assume the variable for total daily allowance

lets say total daily allowance of fat = x grams

Fat consumed at dinner = 48 grams

Fat consumed at dinner in percentage = (Fat consumed at dinner/total daily allowance of fat) × 100

= (48 grams/x grams)×100=(4800/x)%

so (4800/x)%  

So one variable equation that is (4800/x) represents percentage of Emily's dinner fat intake compared to total daily allowance of x gram.

lets take one example

lets says total daily allowance of fat for Emily = 100gm

so from derived equation that is 4800/x , we can get required percentage by putting x =  total daily allowance of fat = 100gm

=4800/100 = 48%.

you can change value of variable x according to total daily allowance and get the required dinner intake percentage by equation 4800/x.


6 0
3 years ago
Aidan has 20 ft of fence with which to build a rectangular dog run. What is the maximum area he can enclose? Enter your answer i
irina1246 [14]
<h2>Maximum area is 25 m²</h2>

Explanation:

Let L be the length and W be the width.

Aidan has 20 ft of fence with which to build a rectangular dog run.

         Fencing = 2L + 2W  = 20 ft

                     L + W = 10

                      W = 10 - L

We need to find what is the largest area that can be enclosed.

     Area = Length x Width

      A = LW

       A = L x (10-L) = 10 L - L²

For maximum area differential is zero

So we have

      dA = 0

      10 - 2 L = 0

        L = 5 m

      W = 10 - 5 = 5 m

Area = 5 x 5 = 25 m²

Maximum area is 25 m²

7 0
3 years ago
What number is ten thousand less than 842719
almond37 [142]
842,719-10,00= 832,719
4 0
3 years ago
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