All categories

4 years ago

For x greater than or equal to zero and less than or equal to 2 pi, sin x and cos x are both decreasing on the intervals

Mathematics

1 answer:

klemol [59]4 years ago

3 0

You made one? Oml.. You could of just texted me. It's B

You might be interested in

Sarah's math teacher said that each question answered correctly on a test would be

Semmy [17]

Hi there
I know it is!

7 0

3 years ago

HELP !! The figure below is made from a rectangle and a triangle use the measurements shown below , find the area of the figure

Vika [28.1K]

The answer is 76. multiply 15 times 4 which is 60.
and the triangle is the difference of the sides, so 19-15=4.
so then multiply the base and height, this leaves 4x4 which is 16.
then add the two. 60+ 16= 76

7 0

4 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 t 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 t = 0 the value of P 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 k 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

4 years ago

How much is 300000 pennies?

DaniilM [7]

If my maths are correct it would be $3,000

7 0

3 years ago

Write the 5th term in the expansion of (5x-y/2)^7

Nezavi [6.7K]

Answer:

65625/4(x^5)(y²)

Step-by-step explanation:

Using binomial expansion

Formula: (n k) (a^k)(b ^(n-k))

Where (n k) represents n combination of k (nCk)

From the question k = 5 (i.e. 5th term)

n = 7 (power of expression)

a = 5x

b = -y/2

....................

Solving nCk

n = 7

k = 5

nCk = 7C5

= 7!/(5!2!) ------ Expand Expression

=7 * 6 * 5! /(5! * 2*1)

= 7*6/2

= 21 ------

.........................

Solving (a^k) (b^(n-k))

a = 5x

b = -y/2

k = 5

n = 7

Substituting these values in the expression

(5x)^5 * (-y/2)^(7-5)

= (3125x^5) * (-y/2)²

= 3125x^5 * y²/4

= (3125x^5)(y²)/4

------------------------------------

Multiplying the two expression above

21 * (3125x^5)(y²)/4

= 65625/4(x^5)(y²)

5 0

3 years ago