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melomori [17]
2 years ago
8

14c^3d^2-21c^2d^3/14cd

Mathematics
1 answer:
levacccp [35]2 years ago
3 0

14c³d²-21c²d³/14cd

= (196c^4d³-21c²d³)/ 14cd

= (196c³d²-21cd²)/14

= (28c³d²-3cd²)/2


I hope that's help !

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An elevator weighing 15,000 newtons is raised one story in 50 seconds. If the distance between stories is 3.0 meters, how much p
Pavel [41]
Power
= Work done  /  Time
= 15000 N * 3 m / 50 s
= 45000/50  N-m/s
= 900 W

It may sound like an extremely low value, but 3.0 / 50 s is also a very slow speed.

8 0
2 years ago
Read 2 more answers
The number of bacteria in a petri dish on the first day was 113 cells. If the number of bacteria increase at a rate of 82% per d
Tresset [83]

Answer:

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

3 0
2 years ago
When Hiroto is writing, there is 0.92 probability that there will be no spelling mistakes on a page. One day, Hiroto writes an e
prohojiy [21]

Answer:

0.60

Step-by-step explanation:

Use binomial probability.

P = nCr pʳ qⁿ⁻ʳ

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

The probability of a mistake on at least one page is 1 minus the probability of making no mistakes.

P(at least 1) = 1 − P(none)

P = 1 − ₁₁C₀ (0.08)⁰ (0.92)¹¹⁻⁰

P = 1 − (0.92)¹¹

P = 0.60

8 0
2 years ago
A population of 550 rabbits is increasing by 7.5% each year. in about how many years will the population be over 1000?
Furkat [3]

I believe this would take the form of an exponential equation:

A = Ao (1 + r)^t

where A is final population, Ao is initial population, r is rate of growth and t is time

 

A / Ao = (1 + r)^t

log A / Ao = t log (1 + r)

t = (log A / Ao) / log (1 + r)

t = [log (1000 / 550)] / log (1.075)

t = 8.27 years

 

SO the answer is B) about 9 years

5 0
2 years ago
Can someone help me with this
Anvisha [2.4K]

Answer:

D. \frac{{7}^{11} }{ {4}^{11} }

Step-by-step explanation:

( \frac{7}{4}) {}^{11} = \frac{7 {}^{11} }{4 {}^{11} }

Since we are multiplying an exponential number to a fraction the answer is going to be

\frac{7 {}^{11} }{4 {}^{11} }

Hope this helps ❤❤❤ ;)

7 0
2 years ago
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