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2 years ago

Solve 3x2 + 4x = 2. ??

Mathematics

1 answer:

Arturiano [62]2 years ago

4 0

Answer: Answer: The answer should be x = -1. Hope this helps!

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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

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

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

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

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

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

8 months ago

Choose the graph below that meets all three of the following requirements:

zmey [24]

Answer:

Your answer would be D

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

ashley spends $12 to buy 3 pencils and 6 rulers. Jack spends $8 to buy 5 pencils and 2 rulers. what is the cost of 2 pencils and

Sveta_85 [38]

Answer:

$ 3 left ashley money

$ 1 left jacks money

$ 8 left ashley money

$ 4 left jacks money

Step-by-step explanation

3 pencils + 6 rulers = 12 - 9 = 3

5 pencils + 2 rulers = 8 - 7 = 1

12 - 4 = 8

8 - 4 = 4

6 0

2 years ago

Read 2 more answers

Question 10

EleoNora [17]

Answer:

a.(-b) ( third choice)

explanation:

-a+b - there is no plus sign for -a.b.

not -a+b

(-a)(-6)=6a (-negative sign and -negative sign=+positive sign)

5 0

2 years ago

From an aeroplane vertically above a straight horizontal plane, the angles of depression of two consecutive kilometres stones on

Airida [17]

You have to build the triangles.

They are such that:
h is the common height
x is the horizontal distance from the plane to one stone
Beta is the angle between x and the hypotenuse

Then in this triangle: tan(beta) = h / x ......(1)

1 - x is the horizontal distance from the plane to the other stone
alfa is the angle between 1 - x and h

Then, in this triangle: tan (alfa) = h / [1 -x ] ...... (2)

from (1) , x = h / tan(beta)

Substitute this value in (2)

tan(alfa) = h / { [ 1 - h / tan(beta)] } =>

{ [ 1 - h / tan(beta) ] } tan(alfa) = h

[tan(beta) - h] tan(alfa) = h*tan(beta)

tan(beta)tan(alfa) - htan(alfa) = htan(beta)

h [tan(alfa) + tan(beta) ] = tan(beta) tan (alfa)

h = tan(beta)*tan(alfa) / (t an(alfa) + tan(beta) )

4 0

2 years ago

Solve 3x2 + 4x = 2. ??​

Solve 3x2 + 4x = 2. ??