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ELEN [110]
3 years ago
13

David loves to read. He is able to read 90 pages in 3 hours and 120 pages in 4 hours.

Mathematics
2 answers:
Lisa [10]3 years ago
8 0
A 30 pages per hour

Because 3•3=9 and add a 0 to 3 30 pages each hour
pshichka [43]3 years ago
4 0
A.
He reads 30 pages per hour.
90/3=30
120/4=30
To for every 1 hour he reads, he is able to read 30 pages.
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Mariana [72]

Answer:

Step-by-step explanation:

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2 years ago
misses Larson made 12 pillows from 16 yards of the same fabric how much fabric was used to make each pillow between what two who
Anna [14]
16 ÷ 12 = 1.33333.....
The answer lies in between the number 1 and 2. 
5 0
3 years ago
The Patronete Winery's tastiest wine must have a 12% alcohol content. How many gallons of wine with a 9% alcohol content must be
erik [133]

Answer: 3000 gallon

Step-by-step explanation:

Here, The total quantity of 15% alcohol content = 3000 gallon

And, in which quantity of alcohol = 15% of 3000 gallon = 450 gallon.

Let the total quantity of 9% alcohol content  = x

In which quantity of alcohol = 9% of x gallon= 9x/100 gallon

Now, according to the question,  The wine with a 9% alcohol content will be mixed with 3,000 gallons of wine with a 15% alcohol content in order to achieve the desired 12% alcohol content.

Thus, the total quantity of mixture of 12% alcohol content = The total quantity of 15% alcohol content + total quantity of 9% alcohol content

= 3000 + x

In which quantity of alcohol = 12 % of ( 3000+x) = (360 + 12x/100) gallon ---(1)

But, the quantity of alcohol in 12 % of alcohol content = quantity of alcohol in 15% of alcohol content +  quantity of alcohol in 9% of alcohol content

The quantity of alcohol in 12 % of alcohol content = (450 + 9x/100) gallon ---(2)

On equating equation (1) and (2)

We get, 360 + 12x/100 = 450 + 9x/100

⇒ 3x/100 = 90

⇒ x = 3000

Therefore, The total quantity of the mixture of 9% alcohol content = 3000


6 0
3 years ago
The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

\frac{K-P(t)}{P(t)} = Ae^{-kt}.

The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

\frac{500-100}{100} = A} and A=4.

So,

\frac{500-P(t)}{P(t)} = 4e^{-0.04t}.

Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

\frac{500-200}{200} = 4e^{-0.04t}.

This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325.

So, the population of rats will be 200 after 25 months.

6 0
3 years ago
What is the square root of 144444444453?
ale4655 [162]

Answer:

380058.475044

3 0
3 years ago
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