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Kitty [74]
3 years ago
7

QF Q2.) Solve the triangle

Mathematics
2 answers:
artcher [175]3 years ago
7 0
60°
11
12

see photo for soln

Whitepunk [10]3 years ago
4 0
C=180-A-B=180-67-53=60deg.

by sine rule

b/sinB=a/sinA
b=a/sinA*sinB
=13/sin(67)*sin(53)
= 11.28

c/sinC=a/sinA
c=a/sinA*sinC
=13/sin(67)*sin(60)
= 12.23

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C. 2 4-ounce's bags and 11 3-ounce's bags

D. 8 4-ounce's bags and 3 3-ounce's bags

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You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then

4x+3y=41.

A. Find all integer solutions:

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  • When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.
  • When x=2, then 3y=33, y=11 - possible.
  • When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.
  • When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.
  • When x=5, then 3y=21, y=7 - possible.
  • When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.
  • When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.
  • When x=8, then 3y=9, y=3 - possible.
  • When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.
  • When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.

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B. 1. 2 + 11 = 13,

2. 5 + 7 = 12,

3. 8 + 3 = 11.

The minimal number of bags is 11.

C. 1. 2·7+11·5=69 cents

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3. 8·7+3·5=71 cents

The cheapest is 1st solution.

D. 1. 2·6+11·5=67 cents

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3. 8·6+3·5=63 cents

The cheapest is 3rd solution.

E. 1. 2·2+11·1.50=$20.50

2. 5·2+7·1.50=$20.50

3. 8·2+3·1.50=$20.50

All solutions offer the same revenue.

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