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goldfiish [28.3K]
3 years ago
13

Which of the following are right triangle congruence theorems

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

Answer:

A C D

Step-by-step explanation:

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A 50 foot rope weighing a total of 32 lbs extended over a cliff that is 35 feet to the ground. A large 8 pound bucket with 19 ga
larisa86 [58]

Answer:

A function that gives the work required in foot-lbs to lift the bucket up x feet from the ground is W=16(50-x)+(19-\frac{x}{5})(8.3)x and  the work to get the bucket to the top of the cliff is 3726 foot-lbs

Step-by-step explanation:

Work done to lift the rope by distance x feet:

W_1=32(\frac{50-x}{2})

Work done to lift the bucket by distance x feet:

W_2=(19-\frac{x}{5})(8.3)x

On reaching top 7 gallons of water spilled out so , on going up by x feet \frac{7x}{35}=\frac{x}{5} gallons of water spilled out.

a function that gives the work required in foot-lbs to lift the bucket up x feet from the ground:

W=16(50-x)+(19-\frac{x}{5})(8.3)x

Now the work to get the bucket to the top of the cliff i.e. x =35

W=16(50-35)+(19-\frac{35}{5})(8.3)(35)

W=3726

Hence, a function that gives the work required in foot-lbs to lift the bucket up x feet from the ground is W=16(50-x)+(19-\frac{x}{5})(8.3)x and  the work to get the bucket to the top of the cliff is 3726 foot-lbs

5 0
3 years ago
A new club sent out 180 coupons to boost sales for next year's memberships. They provided 4 times as many to potential members t
pashok25 [27]

They sent 36 coupons to the existing members

3 0
2 years ago
Please help me out :P
Veseljchak [2.6K]

cos θ = \frac{-4\sqrt{65} }{65}, sin θ = \frac{-7\sqrt{65} }{65}, cot  θ  = 4/7, sec  θ = \frac{-\sqrt{65} }{4}, cosec  θ  = \frac{-\sqrt{65} }{7}

<h3>What are trigonometric ratios?</h3>

Trigonometric Ratios are values of all the trigonometric functions based on the value of the ratio of sides in a right-angled triangle.

Sin θ: Opposite Side to θ/Hypotenuse

Tan θ: Opposite Side/Adjacent Side & Sin θ/Cos

Cos θ: Adjacent Side to θ/Hypotenuse

Sec θ: Hypotenuse/Adjacent Side & 1/cos θ

Analysis:

tan θ = opposite/adjacent = 7/4

opposite = 7, adjacent = 4.

we now look for the hypotenuse of the right angled triangle

hypotenuse = \sqrt{7^{2} + 4^{2} } = \sqrt{49+16} = \sqrt{65}

sin θ = opposite/ hyp = \frac{7}{\sqrt{65} }

Rationalize, \frac{7}{\sqrt{65} } x \frac{\sqrt{65} }{\sqrt{65} } = \frac{7\sqrt{65} }{65}

But θ is in the third quadrant(180 - 270) and in the third quadrant only tan and cot are positive others are negative.

Therefore, sin θ = - \frac{7\sqrt{65} }{65}

cos   θ  = adj/hyp = \frac{4}{\sqrt{65} }

By rationalizing and knowing that cos  θ  is negative, cos θ  = -\frac{-4\sqrt{65} }{65}

cot θ  = 1/tan θ  = 1/7/4 = 4/7

sec θ  = 1/cos θ  = 1/\frac{4}{\sqrt{65} } = -\frac{-\sqrt{65} }{4}

cosec θ  = 1/sin θ  = 1/\frac{\sqrt{65} }{7} = \frac{-\sqrt{65} }{7}

Learn more about trigonometric ratios: brainly.com/question/24349828

#SPJ1

5 0
2 years ago
How do i write this problem
ratelena [41]
3-(s+4) 
less than- subtract
sum- addition 

since it said 3 less than the sum of a number and 4 that means you would have to add s and 4
5 0
3 years ago
Read 2 more answers
98 POINTS!<br> d(e-f)=g<br> Solve For f.
Lubov Fominskaja [6]

d (e - f) = g

de - df = g

df = de - g

f = ed - g/d

Hope helps-Aparri

5 0
3 years ago
Read 2 more answers
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