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cestrela7 [59]
3 years ago
12

PLEASE HELP WITH THE PROBLEM ON THE ATTACHED FILE

Mathematics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

6 : 1

Step-by-step explanation:

b - 3a / 9 = a /3

Cross multiply

3(b - 3a) = 9a

Expand

3b - 9a = 9a

Add 9a to both sides

3b = 9a + 9a

3b = 18a

Divide both sides by 3

b = 6a

Ratio of a : b

6 : 1

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Jeff worked 51 hours last week. How many straight time and overtime hours did he worke?
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B) 40 straight hours - 11 overtime hours

Step-by-step explanation:

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3 years ago
Solving right triangles
Sav [38]
<h2>1. Answer:</h2>

A right triangles is a triangle having a 90 degree side. According to the figure, the sides of this triangle are expressed in inches. Therefore, we can find the missing sides and angles as follows:

<u>m∠B:</u>

The sum of the three interior angles of any triangle is 180°, therefore:

m \angle B + 51^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle B = 39^{\circ}}

<u>CA and AB:</u>

We must use the law of sines as follows:

\frac{CA}{sin39^{\circ}}=\frac{9}{sin51^{\circ}} \\ \\ \therefore CA=\frac{9sin39^{\circ}}{sin51^{\circ}} \\ \\ \therefore \boxed{CA=7.3in}

\frac{AB}{sin90^{\circ}}=\frac{9}{sin51^{\circ}} \\ \\ \therefore AB=\frac{9sin90^{\circ}}{sin51^{\circ}} \\ \\ \therefore \boxed{AB=11.6in}

<h2>2. Answer:</h2>

According to the figure, the sides of this triangle are expressed in meters. Therefore, we can find the missing sides and angles as follows:

<u>m∠A:</u>

The sum of the three interior angles of any triangle is 180°, therefore:

m \angle A + 53^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle A = 37^{\circ}}

<u>CA and CB:</u>

We must use the law of sines as follows:

\frac{CA}{sin53^{\circ}}=\frac{5}{sin90^{\circ}} \\ \\ \therefore CA=\frac{5sin53^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CA=4.0m}

\frac{CB}{sin37^{\circ}}=\frac{5}{sin90^{\circ}} \\ \\ \therefore CB=\frac{5sin37^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CB=3.0m}

<h2>3. Answer:</h2>

According to the figure, the sides of this triangle are expressed in miles. Therefore, we can find the missing sides and angles as follows:

<u>m∠B:</u>

The sum of the three interior angles of any triangle is 180°, therefore:

m \angle A + 28^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle B = 62^{\circ}}

<u>CB and AB:</u>

We must use the law of sines as follows:

\frac{CB}{sin28^{\circ}}=\frac{29.3}{sin62^{\circ}} \\ \\ \therefore CB=\frac{29.3sin28^{\circ}}{sin62^{\circ}} \\ \\ \therefore \boxed{CA=15.6mi}

\frac{AB}{sin90^{\circ}}=\frac{29.3}{sin62^{\circ}} \\ \\ \therefore AB=\frac{29.3sin90^{\circ}}{sin62^{\circ}} \\ \\ \therefore \boxed{AB=33.2mi}

<h2>4. Answer:</h2>

According to the figure, the sides of this triangle are expressed in miles. Therefore, we can find the missing sides and angles as follows:

<u>m∠A:</u>

The sum of the three interior angles of any triangle is 180°, therefore:

m \angle A + 24^{\circ} + 90^{\circ} = 180^{\circ} \\ \\ \boxed{m \angle A = 66^{\circ}}

<u>CA and CB:</u>

We must use the law of sines as follows:

\frac{CA}{sin66^{\circ}}=\frac{14}{sin90^{\circ}} \\ \\ \therefore CA=\frac{14sin66^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CA=12.8mi}

\frac{CB}{sin24^{\circ}}=\frac{14}{sin90^{\circ}} \\ \\ \therefore CB=\frac{14sin24^{\circ}}{sin90^{\circ}} \\ \\ \therefore \boxed{CB=5.7mi}

8 0
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