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Nadya [2.5K]
3 years ago
8

In ΔUVW, w = 11 cm, ∠W=8° and ∠U=132°. Find the area of ΔUVW, to the nearest square centimeter.

Mathematics
2 answers:
Nookie1986 [14]3 years ago
5 0

Answer:

208

Step-by-step explanation:

Delta math

yulyashka [42]3 years ago
4 0
These boys r annoying frfr like stop
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It takes Emma 18 minutes to type and spell check 14 pages of a manuscript. Find how long it takes her to type and spell check 77
NNADVOKAT [17]

Answer:

99 minutes, or an hour and 39 minutes.

Step-by-step explanation:

It takes Emma 18 minutes to do 14 pages. That means that we can set up a proportion, where Emma uses 18 minutes to do 14 pages, which is equal to x minutes per page!

\frac{18}{14} =\frac{x}{1}

14x = 18

7x = 9

x = 9/7

So, each page takes 9/7 minutes for Emma to finish.

She will do 77 pages. 77 * (9/7) = 11 * 9 = 99.

So, it will take Emma 99 minutes, or an hour and 39 minutes, to type and spell check 77 pages.

Hope this helps!

7 0
3 years ago
Read 2 more answers
Describe and correct the error in finding csc θ, given that θ is an acute angle of a right triangle and cos θ =7/11
a_sh-v [17]

Answer:

<h2>cosecθ = 1/sinθ = 11/6√2</h2>

Step-by-step explanation:

Given that  cos θ =7/11, cosec θ = 1/sinθ in trigonometry.

Based on SOH, CAH, TOA;

cosθ = adjacent/hypotenuse = 7/11

adjacent = 7 and hyp = 11

Since sinθ = opp/hyp, we need to get the opposite to be able to calculate sinθ.

Using pythagoras theorem to get the opposite;

hyp^{2} = adj^{2}  + opp ^{2}  \\opp = \sqrt{hyp^{2} - adj^{2}  } \\opp = \sqrt{11^{2} - 7^{2}} \\opp = \sqrt{72} \\opp = 6\sqrt{2}

sinθ = 6√2/11

cosecθ = 1/sinθ = 1/( 6√2/11)

cosecθ = 1/sinθ = 11/6√2

Note the error; cscθ\neq 1/cosθ but cscθ = 1/sinθ

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3 years ago
X -4 ifx&gt;4<br> |<br> Rsm question
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Answer:

x>4

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Step-by-step explanation:

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2 years ago
1. The pollution level in the center of a city at 6 am is 30 parts per million and it
Nady [450]

Answer:

a) y(t) = 30 + 25t

b) The pollution level at 4 o'clock in the afternoon​ is of 280 parts per million.

Step-by-step explanation:

The pollution level in the center of a city at 6 am is 30 parts per million and it grows in linear fashion by 25 parts per million every hour.

This means that the pollution in t hours after 6 am is given by:

y(t) = 30 + 25t, which is the answer to question a.

b. The pollution level at 4 o'clock in the afternoon​

4 in the afternoon is 10 hours after 6 am, so this is y(10).

y(10) = 30 + 25*10 = 30 + 250 = 280

The pollution level at 4 o'clock in the afternoon​ is of 280 parts per million.

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The answer is

4 \frac{2}{3}

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