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Ostrovityanka [42]
4 years ago
7

Find cscx if sinx+cotx cosx= sqrt3

Mathematics
1 answer:
valentinak56 [21]4 years ago
5 0

Answer:

The answer is (d) ⇒ cscx = √3

Step-by-step explanation:

∵ sinx + (cotx)(cosx) = √3

∵ sinx + (cosx/sinx)(cosx) = √3

∴ sinx + cos²x/sinx = √3

∵ cos²x = 1 - sin²x

∴ sinx + (1 - sin²x)/sinx = √3 ⇒ make L.C.M

∴ (sin²x + 1 - sin²x)/sinx = √3

∴ 1/sinx = √3

∵ 1/sinx = cscx

∴ cscx = √3

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Ann [662]
Answer (apply the FOIL method)

X^2+8x+15
6 0
2 years ago
Read 2 more answers
2x3 + 5x2 + 6x + 15
iren [92.7K]

Answer:

C

Step-by-step explanation:

Distribute x^2 to 2x and 5, you’ll get 2x^3+5x^2.

The distribute 3 to 2x and 5, you’ll get 6x+15.

You’ll get what the question is asking for.

7 0
3 years ago
A company is designing a new cylindrical water bottle. The volume of the bottle will be 211 cm^3. The height of the water bottle
Zinaida [17]

Answer:

The correct answer is 2.912 cm.

Step-by-step explanation:

A company is designing a new cylindrical water bottle.

Volume of a cylinder is given by π × r^{2} × h, where h is the height of the cylinder and r is the radius of the cylinder.

The volume of each bottle will be 211 cm^{3}.

The height (h) of the water bottle be 7.9 cm.

Let the radius of the bottle be r cm.

∴ π × r^{2} × h = 211 ; (π = 3.15)

⇒ r^{2} × 24.885 = 211

⇒ r = 2.912

The radius of the water bottle is 2.912 cm.

4 0
3 years ago
Which inequality is equivalent to this one? y-85-2 y-8+82-2+8 oy- 8+8 ​
vladimir2022 [97]

Answer:

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5 0
3 years ago
The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds. Using the empirical ru
coldgirl [10]

If the value of the z-score is 1. Then the probability that a cat will weigh less than 11 pounds will be 0.84134.

<h3>What is the z-score?</h3>

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

The z-score is given as

z = (x - μ) / σ

Where μ is the mean, σ is the standard deviation, and x is the sample.

The weight of a cat is normally distributed with a mean of 9 pounds and a standard deviation of 2 pounds.

Then the probability that a cat will weigh less than 11 pounds will be

The value of z-score will be

z = (11 – 9) / 2

z = 1

Then the probability will be

P(x < 11) = P(z < 1)

P(x < 11) = 0.84134

Thus, the probability that a cat will weigh less than 11 pounds will be 0.84134.

More about the z-score link is given below.

brainly.com/question/15016913

#SPJ1

4 0
2 years ago
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