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2 years ago

What is the estimated weight of three containers weighing 1.656 kg each?

Mathematics

1 answer:

Ede4ka [16]2 years ago

8 0

Answer:

5 or 4.9

i dont know tho hxkw8swlwsaiprjw9sksoqkw

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One more question on this test guys. please answer as fast as possible. If you do,

GaryK [48]

1. The given rectangular equation is $x=2$ .

We substitute $x=r\cos \theta$ .

$r\cos \theta=2$

Divide through by $\cos \theta$

$r=\frac{2}{\cos \theta}$

$r=2}\sec \theta$

$\boxed{x=2\to r=2\sec \theta}$

2. The given rectangular equation is:

$x^2+y^2=36$

This is the same as:

$x^2+y^2=6^2$

We use the relation $r^2=x^2+y^2$

This implies that:

$r^2=6^2$

$\therefore r=6$

$\boxed{x^2+y^2=36\to r=6}$

3. The given rectangular equation is:

$x^2+y^2=2y$

This is the same as:

We use the relation $r^2=x^2+y^2$ and $y=r\sin \theta$

This implies that:

$r^2=2r\sin \theta$

Divide through by r

$r=2\sin \theta$

$\boxed{x^2+y^2=2y\to r=2\sin \theta}$

4. We have $x=\sqrt{3}y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta\sqrt{3}$

This implies that;

$\tan \theta=\frac{\sqrt{3}}{3}$

$\theta=\frac{\pi}{6}$

$\boxed{x=\sqrt{3}y\to \theta=\frac{\pi}{6}}$

5. We have $x=y$

We substitute $y=r\sin \theta$ and $x=r\cos \theta$

$r\cos \theta=r\sin \theta$

This implies that;

$\tan \theta=1$

$\theta=\frac{\pi}{4}$

$\boxed{x=y\to \theta=\frac{\pi}{4}}$

6 0

3 years ago

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

1 year ago

Somebody help me?! Finding an angle measure given a triangle and parallel lines.

balandron [24]

Answer:

x=78

Step-by-step explanation:

the sum of all angles in a triangle=180

180-59=129

y>x

70+59=129

180-129=51

129-51=78

x=78

8 0

3 years ago

The function A = 15x - x2, where x is the width in yards, gives the area A of a llama pen in square yards. Make a table using 1/

Mnenie [13.5K]

Answer:

The width which gives the greatest area is 7.5 yd

Step-by-step explanation:

This is an application of differential calculus. Given the area as a function of the width, we simply need to differentiate the function with respect to x and equate to zero which yields; 15-2x=0 since the slope of the graph is zero at the turning points. Solving for x yields, x=7.5 which indeed maximizes the area of the pen

6 0

3 years ago

9 * 9 ^ 2 x + 3 equal 27 * 3 ^ x - 2

Amiraneli [1.4K]

3^2*3^2 (2x+3)=3^2+4x+6=3^4x+8

3^3*3^x-2=3^x+1

4x+8=x+1

3x=-7

x=-7/3

6 0

3 years ago