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3 years ago

Area of a rectangle 4/5 ft times 2 ft

Mathematics

1 answer:

artcher [175]3 years ago

8 0

Area is found by multiplying length by width. (4/5)(2)=1.6.

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Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

The average annual income, I, in dollars, of a lawyer with an age of x years is modeled with the following function:

Eduardwww [97]

I = -425x^2 + 45500x - 650000
For I to be maximum, dI/dx = 0
dI/dx = -850x + 45500 = 0
x = 45500/850 = 53.53

Therefore, maximum I = -425(53.53)^2 + 45500(53.53) - 650000 = 567,794.11

Therefore, the maximum average annual income, in dollars, a lawyer can earn is $567,794

8 0

3 years ago

Jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

wariber [46]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

The fith term of a G.P is 8 times the 2nd term find its common ratio

Dovator [93]

Answer:

Common ratio r = 2

Step-by-step explanation:

$\because$ the fith term of a G.P is 8 times the 2nd term.

$\therefore \: t_5 = 8\times t_2 \\ \therefore \: ar^{5-1}=8\times \: ar^{2-1} \\ \therefore \: r^{4}=8\times \: r^{1} \\ \therefore \: r^{3}=8 \\ \therefore \:r^{3}= {2}^{3} \\ \hspace{20 pt}\huge \red{ \boxed{\therefore \:r= {2}}} \\$

Hence, common ratio is 2.

6 0

3 years ago

Read 2 more answers

A biologist is studying the growth of a particular species of algae. She writes the fall inclusion to show the radius of the alg

horrorfan [7]

Answer:

Part A)

A reasonable domain to plot the growth function is:

$0\leq d \leq 12$

Part B)

The y-intercept represents that when the biologist started her study, the radius of the algae was five millimeters.

Part C)

The average rate of change from d = 4 to d = 11 was about 0.11. This means that from the 4th day to the 11th, the radius of the algae grew, on average, at a rate of 0.11 mm per day.

Step-by-step explanation:

The radius of the algae f(d) in millimeters after d days is given by the function:

$f(d)=5(1.02)^d$

Part A)

We know that the radius of the algae was approximately 6.34 mm when the biologist concluded her study. To find the reasonable domain, we can substitute 6.34 for f(d) and solve for d. Therefore:

$6.34=5(1.02)^d$

Divide both sides by five:

$(1.02)^d=1.268$

Take the log of both sides with base 1.02:

$\displaystyle d=\log_{1.02}1.268$

Using the Change of Base Property, evaluate for d:

$\displaystyle d=\frac{\log 1.268}{\log 1.02}=11.9903...\approx 12$

So, the biologist concluded her study after 12 days.

Therefore, a reasonable domain to plot the growth function is:

$0\leq d\leq 12$

Part 2)

The y-intercept of the function is when d = 0. Find the y-intercept:

$f(0)=5(1.02)^{(0)}=5(1)=5$

Since d represent the amount of days after the study had begun, the y-intercept represents the radius of the algae on the initial day.

So, when the biologist started her study, the radius of the algae was five millimeters.

Part 3)

To find the average rate of change for a nonlinear function, we find the slope between the two endpoints on the interval.

We want to find the average rate of change of f(d) from d = 4 to d = 11.

Find the endpoints:

$f(4)=5.4121...\text{ and } f(11)=6.2168...$

And find the slope between them:

$\displaystyle m=\frac{f(11)-f(4)}{11-4}=0.1149...\approx 0.11$

Since f(d) measures millimeters and d measures days, this tells us that, on average, the radius of the algae grew by about 0.11 mm per day from the 4th day to the 11th day.

5 0

2 years ago