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uranmaximum [27]
3 years ago
7

Josh has 12 computer games he received 1\4 of them for his birthday how many computer games did he receive for his birthday

Mathematics
1 answer:
katrin2010 [14]3 years ago
5 0

Josh received three. 12/4 or 12*1/4 is 3.

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Mr. Nelson sold 14 bags of popcorn and 21 bottles of water in today. At this rate, how many more bottles of water than bags of p
e-lub [12.9K]

Answer:

He will sell 70 bags of popcorn and 105 bottles

Step-by-step explanation:

In 1 day he sells 14 bags of popcorn and 21 bottles

Multiply 14 by 5 which is 70

then multiply the bottles which is 21 by 5 days

105

He will sell 70 bags of popcorn and 105 bottles

6 0
2 years ago
Read 2 more answers
Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.
Sladkaya [172]

Answer:

The solution is  \frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

Step-by-step explanation:

From the question

    The function given is  f(x) =  \frac{e^{2x}}{ 25 + e^{4x}} dx

The  indefinite integral is  mathematically represented as

          \int\limits  {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx

Now  let  e^{2x} =  u

=>   \frac{du}{dx} 2e^{2x}

=>   2 e^{2x}dx =  du

So

\int\limits  {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx =  \int\limits  {\frac{1}{ 2(25 + u^2)} } \, du

= \frac{1}{2} \int\limits  {\frac{1}{ 25 + u^2)} } \, du

=  \frac{1}{2} \int\limits  {\frac{1}{ 5^2 + u^2)} } \, du

= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5}  +  C

Now substituting for  u

\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] +  C

3 0
3 years ago
Please explain I really need help
velikii [3]

Answer:

1. 1343 years

2. 9 hours

3. 39 years

Step-by-step explanation:

1. Given, half-life of carbon = 5730 years.

∴ λ = 0.693/half-life of carbon = 0.693/5730 = 0.000121

If N₀ = 100 then N = 85

Formula:- N = N₀*e^(-λt)

∴ 85 = 100 * e^(-0.000121t)

∴㏑(-0.85)=-0.000121t

∴ t = 1343 years

2. Given half-life of aspirin = 12 hours

λ = 0.693/12 = 0.5775

Also N₀ = 100 then 70 will disintegrate and N = 30 will remain disintegrated.

∴ 70 = 100 *e^(-0.05775t)

0.70 = e^(-0.05775t)

㏑(0.70) = -0.05775t

∴ t = 9 hours

3.  The population of the birds as as A=A₀*e^(kt)

Given that the population of birds fell from 1400 from 1000, We are asked how much time it will take for the population to drop below 100, let that be x years.

The population is 1400 when f = 0, And it is 1000 when f = 5

We can write the following equation :

1400 = 1000e^(5t).

∴1400/1000 = e^(5k)

∴ k = ㏑(1.4)/5

We need to find x such that 1400/100 = e^(xk)

14 = e^(xk)

∴ x = 39 years

5 0
3 years ago
Need help with this! Due in 10 mins please
svetlana [45]
True for some cases
7 0
3 years ago
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Enter the equation of the line in slope-intercept form.
antiseptic1488 [7]

Answer:

ajfbgaulfgsuidglml

Step-by-step example:

4 0
3 years ago
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