From two points on the same level as the base of a tree, the angles of elevation to the top of the tree are found to be 24° and 46°. If the two points are 42 feet apart and on the same side of the tree, how tall is the tree? (Give your answer to the nearest tenth of an inch.)​

Answers

Answer 1
Answer:

Answer:

the answer is 32.8in.

Step-by-step explanation:

(sin 22)/42 = (sin 24)/x

x = 45.60

sin 46 = x/45.60

x = 32.80

32.8in.

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A 40 N force is applied to an object with a mass of 0.1 kg on a frictionless surface. What is the acceleration of the object?
Add the following - 4/9,7/12and - 3/8​
Write 2 equivalent ratios for each ratio given:2/73:3 1/22.5 to 6

-5 1/2 + 6 3/4 + (-4 1/4)

Answers

Answer:

-5(1)/(2) +6(3)/(4)+(-4(1)/(4))

-3 is the answer

Simplify 8X +7 - 2X -4

Answers

Answer:

6x + 3

Step-by-step explanation:

8x - 2x + 7 -4

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Write the phrase as an expression. Then evaluate when x=5 .6 more than the product of 8 and a number x

Expression:

When x=5, the value of the expression is

Answer:

Answers

Answer:

The answer is 46 when x = 5

Step-by-step explanation:

8*x + 6

So x = 5

8*5 + 6

40 + 6

46

Answer:

b

Step-by-step explanation:

For four weeks in June Cameron baked 3 1/4 miles each week and swim 2 1/2 miles each week for three weeks in July he baked 4 3/4 miles each week and swim 3 1/2 miles each week how much greater was the total distance Cameron bike and swim in July compared to the total distance he bike in swim in June

Answers

For June:
 Bike:
 (4) * (3 1/4) = 13 miles
 Swim:
 (4) * (2 1/2) = 10 miles
 Total:
 13 + 10 = 23 miles
 For July:
 Bike:
 (3) * (4 3/4) = 14.25 miles
 Swim:
 (3) * (3 1/2) = 10.5 miles
 Total:
 14.25 + 10.5 = 24.75 miles
 The difference between both months is:
 24.75 - 23 = 1.75 miles
 Answer:
 
the total distance Cameron bike and swim in July compared to the total distance he bike in swim in June was 1.75 miles greater

What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?

Answers

Completed question:

An initial time study resulted in an average observed time of 2.2 minutes per cycle, and a standard deviation of .3 minutes per cycle. The performance rating was 1.20. What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?

Answer:

47

Step-by-step explanation:

When doing a statistic study, a sample of the total amount must be taken. This sample must be done randomly, and, to be successful, the sample size (n) must be determined, by:

n = ((Z_(\alpha/2)*S )/(E))^2

Where Z(α/2) is the value of the standard normal variable associated with the confidence, S is the standard deviation, and E is the precision. The confidence indicates if the study would have the same result if it would be done several times. For a confidence of 95.44, Z(α/2) = 2.

The standard deviation indicates how much of the products deviate from the ideal value, and the precision indicates how much the result can deviate from the ideal. So, if it may vary 4% of the true value (2.2), thus E = 0.04*2.2 = 0.088.

n = [(2*0.3)/0.088]²

n = 46.48

n = 47 observations.

Warehouse Club A charges its members $55 to join plus $25 Warehouse Club B charges a $10 to join plus $40 each month.

Answers

Warehouse Club A: y= $25x + $55
Warehouse Club B: y= $40x + $10