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Natasha2012 [34]
3 years ago
8

One crate can hold 8 cases of trading cards.How many crates are needed to hold 128 cases of trading cards?

Mathematics
1 answer:
Lunna [17]3 years ago
7 0
16 crates to hold all 128  cases of trading cards

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Anyone one know the answer?
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Read 2 more answers
Members of a college chemistry department agree to contribute equal amounts of money to make up a scholarship fund of $280. Then
Zarrin [17]

Answer:

7 members

Step-by-step explanation:

Lets say that the department has n members. If each member contributes with an equal amount x and make a total of $280, we know that:

n*x = 280

Where each one colaborated with:

n*x/n = 280/n

Or what is the same that: (as the n on the left eliminate themselves)

x = 280/n (*)

Now the department hires 3 more members, so we now have n+3 members. And we know that the pass from contributing x to contribute $30 less, it is, x-30 (I omit the $ symbol for simplicity). So, know we have:

x-30 = 280/(n+3)

We can replace x here by the formulation of x we did in the (*) equation:

( 280/n) - 30 = 280/(n+3)

using common denominator:

( 280/n) - 30 n/n = 280/(n+3)

( 280 - 30 n)/n = 280/(n+3)

Cross multiplying the denominators:

( 280 - 30 n)*(n+3) = 280*(n)

280n - 30n^2 + 840 - 90n = 280n

Subtracting 280 n in both sides:

30n^2 + 840 - 90n = 0

Dividing both sides by 30:

-n^2 - 3n + 28 = 0

Factorizing this term (you can notice the factor immediately or solving with baskara resolution, I omit the baskara for simplicity):

-(n + 7)(n - 4) =0

So, n is equal to 4 and -7. As we are dealing with people we can not have -7 people, so we keep only n=4.

So, previous the new hiring there were 4 members, now there are 7.

The used to collaborate with 280/4 = $70, now the collaborate with 280/7=$40, and we see the collaboration decreased in $30.

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Write the number seven hundred fifty -one with digits
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Solve the following system
scZoUnD [109]

Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

4 0
3 years ago
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