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barxatty [35]
2 years ago
10

The angle of depression from the top of the school to the base of the flag pole in front of the school is 50° . If the flag pole

is 35 feet from the base of the school, find the height of the school.
A) 38.4 ft B) 39.2 ft C) 39.8 ft D) 40.5 ft E) 41.7 ft
Mathematics
1 answer:
Vinvika [58]2 years ago
4 0

Answer:

41.7ft

Step-by-step explanation:

Tan50°=h/35

35×tan50=h

h=41.71

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Answer:

<em>75%</em>

Step-by-step explanation:

27 ÷ 36 = 0.75

0.75 → 75%

75% of the crew are men

Hope this helped! :)

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The hourly cost of renting office space in Manhattan is shown by the graph below. The hourly cost of renting office space in Bro
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Will this happen in avenger 4?
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In chess, the Endgame is where you sacrafice pawns, or in this case, minor characters in order to get a powerful piece back. You can leave some pawns in battle while regaining power pieces. The pawns sacraficed were Peter, Stephen, Bucky, Drax, T'challa. Mantis, etc. While leaving behind two pawns: Nebula, and Bruce Banner. Normally I like to think of the Hulk as a Rook, but since he's completely useless at the moment, he's a pawn. Nebula's fairly worthless, so she's a pawn. Thanos is playing with all power pieces and one pawn: Gamora. He sacraficed his pawn in order to complete his queen equivilence: the gauntlet. Now he's playing with all power pieces, while the Avenger's have sacraficed their pawns in order to get their queen: Captain Marvel, who in turn will wage war on Thanos only to find that a pawn has made it across the board and turned into the Hulk, and fights side by side the original Avengers to get the soul stone, revive Tony, who probably dies, get their friends home, welcome new friends, and kill Thanos.

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4 0
3 years ago
If you rotate the point (2, 3) 90 degrees counterclockwise about the origin, what is the image of the point?
AlexFokin [52]
Rotation counterclockwise 90 degrees has the rule

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4 0
3 years ago
Would appreciate the help ! ​
aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 2

\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\

Part 3

\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity \sin^2(\theta)+\cos^2(\theta) = 1 in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0
2 years ago
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