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mojhsa [17]
2 years ago
9

Solve the equation 49k^2 = 64 by factoring.Can anyone help me pls?​

Mathematics
1 answer:
adoni [48]2 years ago
4 0

Answer:

value of

k =  \frac{8}{7}  \: or \: 1.14

Step-by-step explanation:

here's the solution ; -

=》

49k {}^{2}  = 64

=》

(7k) {}^{2}  =  {8}^{2}

=》

\sqrt{ {(7k)}^{2} }  =  \sqrt{ {8}^{2} }

=》

7k = 8

=》

k =  \frac{8}{7}

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Write three numbers greater than 20 that have 9 as a factor
Lubov Fominskaja [6]
We just need to find the multiples of 9 that are greater than 20.

9*1 = 9
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9*3 = 27 
9*4 = 36
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6 0
3 years ago
Need help on this problem
faust18 [17]

Using an exponential function, it is found that:

  • For Country A, the doubling time is of 43 years.
  • For Country B, the growth rate is of 1.9% per year.

<h3>What is the exponential function for population growth?</h3>

The exponential function for population growth is given as follows:

P(t) = P(0)e^{kt}

In which:

  • P(t) is the population after t years.
  • P(0) is the initial population.
  • k is the exponential growth rate, as a decimal.
  • t is the time in years.

For Country A, we have that k = 0.016. The doubling time is t for which P(t) = 2P(0), hence:

P(t) = P(0)e^{kt}

2P(0) = P(0)e^{0.016t}

e^{0.016t} = 2

\ln{e^{0.016t}} = \ln{2}

0.016t = \ln{2}

t = \frac{\ln{2}}{0.016}

t = 43 years.

For Country B, P(36) = 2P(0), hence we have to solve for k to find the growth rate.

P(t) = P(0)e^{kt}

2P(0) = P(0)e^{36k}

e^{36k} = 2

\ln{e^{36k}} = \ln{2}

36k = \ln{2}

k = \frac{\ln{2}}{36}

k = 0.019.

For Country B, the growth rate is of 1.9% per year.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

3 0
1 year ago
I will give you brainliest whoever answers correct first!!
Firlakuza [10]

Answer:

74.4 (llllllllllll ignore this i need 20 characters to answer)

3 0
3 years ago
Please help with this problem ​
Svetlanka [38]
The answer should be 25.64$
You multiple that by 8% and then what you get out of it you add on to what you got as the total .
6 0
3 years ago
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