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Wittaler [7]
3 years ago
5

CosX sec^2X tanX - cosX tan^3X = sinX

Mathematics
1 answer:
ch4aika [34]3 years ago
4 0
\cos x\sec^2x\tan x-\cos x\tan^3x=\cos x\tan x\left(\sec^2x-\tan^2x\right)=\sin x\left(\sec^2x-(\sec^2x-1)\right)=\sin x
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2. A faucet for a 30-gallon bathtub fills at a rate of 3 gallons per minute.
denis-greek [22]

Answer:

20

Step-by-step explanation:

Knowing that it is a 30-gallon bathtub.

And it fills at a rate of 3 gallons per minute.

But also drains 1.5 gallons per minute.

So this is for the first minute.

3 - 1.5 = 1.5

So there will be 1.5 gallons of water in the tub after the first minute.

Now do,

30 ÷ 1.5 = 20

It will take 20 full minutes for the bathtub to fill up.

6 0
3 years ago
Can someone help me with this question please?
Anna71 [15]

Answer:

  36 ft²

Step-by-step explanation:

If BE = 3√2 ft, then the diagonal BD is 6√2 ft. The area of the square (as with any rhombus) is half the product of the diagonals. Since both diagonals are the same length, the area of the square is ...

  area = 1/2·(6√2 ft)² = 36 ft²

The area of the square is 36 square feet.

_____

You may be expected to find the length of one side of the square using the Pythagorean theorem.

  AB² = AE² + EB² . . . . . = the area of the square

  AB² = (3√2 ft)² +(3√2 ft)² = 18 ft² + 18 ft²

  AB² = 36 ft² . . . . . . . . . the area of the square

8 0
3 years ago
The drawing of a model race car shows a length of 2 inches. The length of the actual model car is 16 inches.
erica [24]

Answer:

oc

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white
Rom4ik [11]

Answer:

Part a: <em>The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b: <em>The case in such a way that the chances are maximized so the case  where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c: <em>The minimum and maximum probabilities of winning  for n number of balls are  such that </em>

  • <em>when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n</em>
  • <em>when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5</em>

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

so

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

so

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball  and 1 white ball in urn A and 2 white balls are in urn A'

so

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball  and 2 white balls in urn A and 1 white ball are in urn A'

so

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

<em>As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.</em>

Part b:

<em>As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.</em>

Part c:

The minimum and maximum probabilities of winning  for n number of balls are  such that

  • when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

  • when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0
3 years ago
What is 5 (p-10) expanded​
V125BC [204]
Answer:

5p - 50

You distribute

5 x p
5 x -10
3 0
2 years ago
Read 2 more answers
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