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ivanzaharov [21]
1 year ago
12

Need help with this problem.

Mathematics
1 answer:
drek231 [11]1 year ago
6 0

\cfrac{6^3\cdot 15^3}{(7^0)^3}\implies \cfrac{(6\cdot 15)^3}{(1)^3}\implies \cfrac{90^3}{1}\implies 729000

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You need to arrange 10 of your favorite books on a small shelf. How many different ways can you arrange the books, assyming that
tensa zangetsu [6.8K]

Answer:

You can arrange the books 3,628,800 different ways.

Step-by-step explanation:

You can imagine that the shelf has 10 "slots" where you can place a book. Each slot can take 1 of the 10 books. However, after you have placed 1 book, there are only 9 possible books to place. After 2 have been placed, there are only 8 possible.

Going left-to-right, the first slot has 10 available books.

Total combinations: 10

The 2nd slot has 9 available books.

Total combinations: 10 * 9

The 3rd slot has 8 available books.

Total combinations: 10 * 9 * 8

This continues for all 10 slots.

Eventually, for slot 10, there is only 1 book possible.

Total combinations: 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Multiply all numbers between 10 and 1 with each other is also known as "10 factorial" or "10!"

Thus, the total number of combinations is:

10! = 10*9*8*7*6*5*4*3*2*1 = 3628800

Answer: You can arrange the books 3,628,800 different ways.

8 0
3 years ago
NEED ANSWERED ASAP!
svetlana [45]

Answer:

a- reflection across y axis

b-rotation

c- reflect over y then x axis

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
I need help ASAP pleaseeeeeeee
pav-90 [236]

Answer:

1 / 5^11

Step-by-step explanation:

5^ -6  * 5 ^ -5

We know that a^ b * a^ c = a^( b+c)

5^ ( -5+-5)

5^ -11

We also know that

a^-b = 1/ a^b

1 / 5^11

8 0
2 years ago
Read 2 more answers
(1 point) A bucket that weighs 3.6 pounds and a rope of negligible weight are used to draw water from a well that is 78 feet dee
Deffense [45]

Answer:

The total work done in pulling the bucket to the top of the well is approximately 3,139.1 ft·lb

Step-by-step explanation:

The given parameters are;

The mass of the bucket, W = 3.6 pounds

The depth of the well, h = 78 feet deep

The mass of water in the bucket = 38 ponds

The rate at which the water is pulled up = 2.9 feet per second

The rate at which water is leaking from the bucket, \dot m = 0.1 pounds per second

We separate and find the work done for lifting the bucket and the water individually, then we add the answers to get the solution to the question as follows;

The work done in lifting bucket empty from the well bottom, W_b = W × h

∴  W_b = 3.6 pounds × 78 feet = 280.8 ft-lb

The work done in lifting bucket empty from the well bottom, W_b = 280.8 ft-lb

The time it takes to lift the bucket from the well bottom to the top, 't', is given as follows;

Time, t = Distance/Velocity

The time it takes to pull the bucket from the well bottom is therefore;

t = 78 ft./(2.9 ft./s) ≈ 26.897

The time it takes to pull the bucket from the well bottom to the top, t ≈ 26.897 s

The mass of water that leaks out from the bucket before it gets to the top, m₂, is therefore;

m₂ = \dot m × t

∴ m₂ = 0.1 lbs/s × 26.897 s = 2.6897

The mass of the water that leaks, m₂ = 2.6897 lbs

The mass of water that gets to the surface m₃ = m - m₂

∴ m₃ = 38 lbs  - 2.6897 lbs ≈ 35.3103 lbs

Given that the water leaks at a constant rate the equation representing the mass of the water as it is lifted can b represented by a straight line with slope, 'm' given as follows;

The slope of the linear equation m = (38 lbs - 35.3103 lbs)/(78 ft. - 0 ft.) = 0.03448\overline 3 lbs/ft.

Therefore, the equation for the weight of the water 'w' can be expressed as follows;

w = 0.03448\overline 3·y + c

At the top of the well, y = 0 and w = 38

∴ 35.3103 = 0..03448\overline 3 × 0 + c

c = 35.3103

∴ w = 0.03448\overline 3·y + 35.3103

The work done in lifting the water through a small distance, dy is given as follows;

(0.03448\overline 3·y + 38) × dy

The work done in lifting the water from the bottom to the top of the well, W_{water}, is given as follows;

W_{water} = \int\limits^{78}_0 {0.03448\overline 3 \cdot y + 35.3103 } \, dy

\therefore W_{water} = \left [ {\dfrac{0.03448\overline 3 \cdot y^2}{2}   + 35.3103 \cdot y\right ]^{78}_0

W_{water} = (0.034483/2 × 78^2 + 35.3103 × 78) - (0.034483 × 0 + 38 × 0)  ≈ 2,859.1

The work done in lifting only the water, W_{water} ≈ 2,859.1 ft-lb

The total work done, in pulling the bucket to the top of the well, W = W_b + W_{water}

∴ W = 2,859.1 ft.·lb + 280.8 ft.·lb ≈ 3,139.1 ft·lb

The total work done, in pulling the bucket to the top of the well, W ≈ 3,139.1 ft·lb.

6 0
2 years ago
The work that Yi did to find the greatest common factor of 42 and 63 is shown below. Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42 Fa
marishachu [46]
Yi left a factor of 63, that is 21.
So the GCF will be 21
5 0
3 years ago
Read 2 more answers
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