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RUDIKE [14]
3 years ago
12

The square root of 7 is between which of the following pair of number?

Mathematics
2 answers:
mixer [17]3 years ago
5 0

Answer:

A

Step-by-step explanation:

because the square root of 4 is 2

and the square root of 9 is 3

since 7 is in between 4 and 9 the square root is in between 2 and 3

enyata [817]3 years ago
4 0

Answer:

A. 2 and 3

Step-by-step explanation:

The square root of 7 is 2.7 which makes it between 2 and 3.

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37.5 feet.

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Factorise this expression as fully as you can 12x^3-9x^2
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7 0
3 years ago
Suppose that the population​ P(t) of a country satisfies the differential equation dP/dt = kP (600 - P) with k constant. Its pop
jeka94

Answer:

The country's population for the year 2030 is 368.8 million.

Step-by-step explanation:

The differential equation is:

\frac{dP}{dt}=kP(600 - P)\\\frac{dP}{P(600 - P)} =kdt

Integrate the differential equation to determine the equation of P in terms of <em>t</em> as follows:

\int\limits {\frac{1}{P(600-P)} } \, dP =k\int\limits {1} \, dt \\(\frac{1}{600} )[(\int\limits {\frac{1}{P} } \, dP) - (\int\limits {\frac{}{600-P} } \, dP)]=k\int\limits {1} \, dt\\\ln P-\ln (600-P)=600kt+C\\\ln (\frac{P}{600-P} )=600kt+C\\\frac{P}{600-P} = Ce^{600kt}

At <em>t</em> = 0 the value of <em>P</em> is 300 million.

Determine the value of constant C as follows:

\frac{P}{600-P} = Ce^{600kt}\\\frac{300}{600-300}=Ce^{600\times0\times k}\\\frac{1}{300} =C\times1\\C=\frac{1}{300}

It is provided that the population growth rate is 1 million per year.

Then for the year 1961, the population is: P (1) = 301

Then \frac{dP}{dt}=1.

Determine <em>k</em> as follows:

\frac{dP}{dt}=kP(600 - P)\\1=k\times300(600-300)\\k=\frac{1}{90000}

For the year 2030, P (2030) = P (70).

Determine the value of P (70) as follows:

\frac{P(70)}{600-P(70)} = \frac{1}{300} e^{\frac{600\times 70}{90000}}\\\frac{P(70)}{600-P(70)} =1.595\\P(70)=957-1.595P(70)\\2.595P(70)=957\\P(70)=368.786

Thus, the country's population for the year 2030 is 368.8 million.

3 0
3 years ago
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