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KiRa [710]
3 years ago
10

Dan has bags of candy with two pieces in each. Bill has bags of candy with five pieces in each. Bill has six more bags than Dan.

The total number of pieces candy the two boys have 100. How many bags does Bill have?
Mathematics
1 answer:
Anna35 [415]3 years ago
7 0

<u>Answer:</u>

Bill has bags of candy with five pieces in each. The Bill has 16 bags

<u>Solution:</u>

Given, Dan has bags of candy with two pieces in each.  

Bill has bags of candy with five pieces in each.  

Bill has six more bags than Dan.  

Then, number of bags with bill = 6 + number of bags with dan.

The total number of pieces candy the two boys have 100.  

Now, candies with bill + candies with dan = 100

⇒ 5 pieces per bag x number of bags with bill + 2 pieces per bag x number of bags with dan = 100

⇒ 5 x (6 + number of bags with dan) + 2 x number of bags with dan = 100

⇒ 30 + 5 x number of bags with dan + 2 x number of bags with dan = 100

⇒ (5  + 2) x number of bags with dan = 100 – 30

⇒ 7 x number of bags with dan = 70

⇒ Number of bags with dan = 10

So, number of bags with bill = 6 + 10 = 16

Hence, bill has 16 bags.

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The Colonel spots a campfire at a bearing N 59∘59∘ E from his current position. Sarge, who is positioned 242 feet due east of th
Rashid [163]

Answer:

i. Colonel is about 201 feet away from the fire.

ii. Sarge is about 125 feet away from the fire.

Step-by-step explanation:

Let the Colonel's location be represented by A, the Sarge's by B and that of campfire by C.

The total angle at the campfire from both the Colonel and Sarge = 59^{0} + 34^{0}

                                           = 93^{0}

Thus,

<CAB = 90^{0} - 59^{0} = 31^{0}

<CBA = 90^{0} - 34^{0} = 56^{0}

Sine rule states;

\frac{a}{Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}

i. Colonel's distance from the campfire (b), can be determined by applying the sine rule;

\frac{b}{Sin B} = \frac{c}{Sin C}

\frac{b}{Sin 56^{0} } = \frac{242}{Sin 93^{0} }

\frac{b}{0.8290} = \frac{242}{0.9986}

cross multiply,

b = \frac{0.8290*242}{0.9986}

  = 200.8993

Colonel is about 201 feet away from the fire.

ii. Sarge's distance from the campfire (a), can be determined by applying the sine rule;

\frac{a}{Sin A} = \frac{c}{Sin C}

\frac{a}{Sin 31^{0} } = \frac{242}{Sin 93^{0} }

\frac{a}{0.5150} = \frac{242}{0.9986}

cross multiply,

a = \frac{0.5150*242}{0.9986}

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Sarge is about 125 feet away from the fire.

8 0
3 years ago
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There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area. 
only A and E satisfy both inequality, in the overlapping shaded area.

 
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true. 
 for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
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