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vaieri [72.5K]
3 years ago
6

Find the slope of the line whose equation is 8y = 2x 4.

Mathematics
1 answer:
Jobisdone [24]3 years ago
4 0
- 8y = 2x+4 - y = (2x+4)/8 - y = 2(x+2)/8 - simplifies to (x+2)/4 - which can be rewritten as y = (x/4)+(2/4) - which is also equivalent to y = 0.25x + 0.5 if y = mx+b is your standard equation, m represents slope thus 0.25 (or 1/4) is your slope value
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Calculus 3 help please.​
Reptile [31]

I assume each path C is oriented positively/counterclockwise.

(a) Parameterize C by

\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}

with -\frac\pi2\le t\le\frac\pi2. Then the line element is

ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt

and the integral reduces to

\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt

The integrand is symmetric about t=0, so

\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt

Substitute u=\sin(t) and du=\cos(t)\,dt. Then we get

\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}

(b) Parameterize C by

\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{3^2+4^2} \, dt = 5\,dt

and

\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt

Integrate by parts with

u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}

\displaystyle \int u\,dv = uv - \int v\,du

\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}

(c) Parameterize C by

\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}

with 0\le t\le1. Then

ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt

and

\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}

8 0
1 year ago
Please help me with number 1 to 6 please
Ugo [173]
1.) 144
2.) 8
3.) 262.5
4.) 36
5.) 630
6.) 440
8 0
2 years ago
Which relation is a function?
puteri [66]

let's recall the vertical line test, if a vertical line hits the graph or points twice, then is NOT a function.  Check the picture below.

6 0
1 year ago
Read 2 more answers
HELP<br><br><br><br><br> QUICK<br><br><br><br><br> LOL THANKS !!!
Anon25 [30]

Answer:

0.6

Step-by-step explanation:

The x increases by 5, while y increases by 3

Lets use the method, which is \frac{rise}{run}.

Keep in mind that x goes horizontally along the axis, while y goes vertically on the axis.

So plug the numbers in.

\frac{3}{5} = 0.6

Hence, the constant proportionality is 0.6.

8 0
3 years ago
Read 2 more answers
Ik it’s pretty easy but u knowh am not really sure if am right or not
gladu [14]

2,6,18

r = 6/2 = 3

18 *3 = 54

54*3 = 162

162*3=486

Answer: 54,162,486

8 0
3 years ago
Read 2 more answers
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