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Nostrana [21]
3 years ago
11

Team A is located on the south bank of a river.Team B and C are located on the north bank.The angle formed from Team C to Team A

is S 42 W.Team C is 10 feet due east of Team B and 12 feet from Team A.How far is Team A from Team B?

Mathematics
1 answer:
kotegsom [21]3 years ago
6 0

Answer:

9.1 feet

Step-by-step explanation:

Please refer to the image attached for explanations

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a set of measuring cups has measures of 1 cup, 3/4 cup, 1/2 cup, 1/3 cup, and 1/4 cup. How could I measure 1/6 of a cup using th
Sophie [7]

1/2 cup = 3/6 cup

1/3 cup =2/6 cup

Just dump a much from the 1/2 cup into the 1/3 cup.

The 1/2 cup will have 1/6 left.

7 0
3 years ago
A supermarket sells corn flakes for $3.60 per box. If the markup on the cereal is 30%, how much did the supermarket pay for each
Alexxx [7]

Answer: 70

Step-by-step explanation:

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Who knows how to do this please i need help I don’t understand
ArbitrLikvidat [17]

Answer:

9,4,0,1

Step-by-step explanation:

They give you different values for the X so all you have to do is plug in the X to the different equations

3 0
3 years ago
Select all ratios equivalent to <br> 22:10.<br><br> 11:5<br> 110:50<br> 33:15
saveliy_v [14]

Answer:

All the 3 choices are correct.

Step-by-step explanation:

11:5   11 divided by 22 is 1/2  and 5/10 is 1/2

110/50   110 divided by 22 is 5 and 50 divided by 10 is 5

33/15     33 divided by 22 is 1 1/2 and 15 divided by 10 is 1 1/2

5 0
3 years ago
Read 2 more answers
In ΔABC, the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively.
Irina-Kira [14]
Given the values of the three sides of the triangle, we can apply the Cosine Law to find the angles of the triangle. Recall that for we can express the value of c through the equation below.

c^{2} = a^{2} + b^{2} - 2abcosC

Rearranging this equation, we can find the value ∠C as shown below.

\cos C = \frac{a^{2}+b^{2}-c^{2}}{2ab}
C = cos^{-1} (\frac{a^{2}+b^{2}-c^{2}}{2ab})

We can apply the same reasoning for finding the value of ∠B as shown.

B = cos^{-1} (\frac{a^{2}+c^{2}-b^{2}}{2ac})

Plugging in the values of the sides (see image attached) from the given. It will now be straightforward to compute for ∠B and ∠C.

C = cos^{-1} (\frac{22.5^{2}+18^{2}-13.6^{2}}{2(22.5)(18)})
C \approx 37.19

B = cos^{-1} (\frac{22.5^{2}+13.6^{2}-18^{2}}{2(22.5)(13.6)})
B \approx 53.13

Answer: ∠C = 37.19° and ∠B = 53.13°

7 0
3 years ago
Read 2 more answers
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