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Elis [28]
3 years ago
5

Michael, Leo and JuWoong collect stamps. The ratio of the number of Michael's stamps to JuWoong's stamps is 2:7. The number of J

uWoon's stamps is 3/4 of the number of Leo's stamps. If Leo has 396 more stamps than Michael, how many stamps do they have altogether?
Mathematics
1 answer:
Alisiya [41]3 years ago
4 0

Answer:

990

Step-by-step explanation:

M/J = 2/7

J = 3/4 L

L = M + 396

7M = 2J

J = 3.5M

3.5M = 3/4 L

L = 4/3(3.5M)

L = 14/3 M

14/3 M = M + 396

11/3 M = 396

M = 3/11 × 396

M = 108

L = M + 396

L = 108 + 396

L = 504

J = 3/4 L

J = 3/4 × 504

J = 378

M + L + J = 108 + 504 + 378 = 990

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Answer:

1 N 3

Step-by-step explanation:

8 0
3 years ago
A metal strip of 44 inches long is to be cut in 2 pieces. the smaller piece to be 12 inches shorter than the larger piece. find
Ket [755]

Answer:

Shorter Piece is 16 inches and Longer Piece is 28 Inches.

Step-by-step explanation:

Cut the 44 Inch piece in half (22+22) and subtracted 6 from one end and added six to another end which is 12 inches in total.

Hope this helps.

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3 0
2 years ago
The Johnsons framed a family picture to hang on the wall. The perimeter of the Frame is 72 inches. Find the length of the frame
olganol [36]
14+14+x+x=72
28+2x=72
2x=44
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7 0
2 years ago
Solve the system algebraically. Check your work. 5x + 2y = 10 3x + 2y = 6 Make sure there are NO SPACES in your answer. Be sure
e-lub [12.9K]
For the first equation its y=5-2/5x or y=-2/5x+5
5 0
3 years ago
Read 2 more answers
How many different 5-person subcommittees can be formed from a club having 12 members
Wittaler [7]
Given that there are 12 persons, the first choice may be in 12 different ways, the second choice may be in 11 different ways, ther third in 10 different ways, the fourth in 9 different ways and the fith in 8 different ways, for a total of:

12x11x10x9x8 different combinations.

Now you have to take in account that 5x4x3x2 are repetitions. So you have to divide the previos counting by 5x4x3x2.

(12x11x10x9x8)/(5x4x3x2) = 792 different subcommittees.

Also, you can use the formula for combinations: C(m,n) = m! / (n! (m-n)!)

C (12, 5) = 12! / (5!) (12-5)! = [12x11x10x9x8x7!] / [5! 7!] = [12x11x10x9x8]/[5x4x3x2] = 792
3 0
2 years ago
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