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3 years ago

A point has no length or width true or false

2 answers:

8 0

The answer should be true. Because the point it has a line with length, and there is no width or thickness, so it would be have one dimensional. Hope it helped you! Have a great day!

-Charlie

serg [7]3 years ago

4 0

The answer would be true. a point is just a single place, and can define no length or width

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(Look at picture above) Find the volume of the cone. Round your answer to the nearest tenth. 9 yd 7 yd The volume of the cone is

nata0808 [166]

Answer:

V = 115.5 cubic yards

Step-by-step explanation:

V (cone) = 1/3πr²h

just substitute given values to get:

r = 7/2 or 3.5

h = 9

V = π(3.5²)(9)/3

V = 115.5 cubic yards

3 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.

If that's the case, then

$z_s=z_xx_s+z_yy_s$

$z_t=z_xx_t+z_yy_t$

as the hint suggests. We have

$z=\sin x\cos y\implies\begin{cases}z_x=\cos x\cos y=\cos(s+t)\cos(s^2t)\\z_y=-\sin x\sin y=-\sin(s+t)\sin(s^2t)\end{cases}$

$x=s+t\implies x_s=x_t=1$

$y=s^2t\implies\begin{cases}y_s=2st\\y_t=s^2\end{cases}$

Putting everything together, we get

$z_s=\cos(s+t)\cos(s^2t)-2st\sin(s+t)\sin(s^2t)$

$z_t=\cos(s+t)\cos(s^2t)-s^2\sin(s+t)\sin(s^2t)$

8 0

3 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

Jan received -22 points on her exam. She got 11 questions wrong out of 50 questions. How much was Jan penalized for each wrong a

svp [43]

2 points for each incorrect answer.

6 0

3 years ago

Read 2 more answers

What is this please anyone

xz_007 [3.2K]

The answer is 1.4 x 10^3 Or 1400
Rewrite the equation as
(2x7) x (10^6 x 10^4) then calculate from there to get your answer

4 0

3 years ago