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Gnoma [55]
3 years ago
15

Find the volume of the triangle

Mathematics
1 answer:
PolarNik [594]3 years ago
4 0

Answer:To calculate the volume of a triangular prism, measure the width and height of a triangular base, then multiply the base by the height by 1/2 to determine the triangle's area. Next, measure the height of the triangular prism and multiply this by the triangle's area to get the volume.

Step-by-step explanation:

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Maria is m inches tall. Blaire is 2 inches taller than Maria. If Blaire is 52 inches tall, which equation could be used to find
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The answer would be C. m+2=52
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I can't find area of the triangle
Andreas93 [3]

Answer:

First do the Base multiplied by the Height.

After divide by 2

Step-by-step explanation:


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interpret r(t) as the position of a moving object at time t. Find the curvature of the path and determine thetangential and norm
Igoryamba

Answer:

The curvature is \kappa=1

The tangential component of acceleration is a_{\boldsymbol{T}}=0

The normal component of acceleration is a_{\boldsymbol{N}}=1 (2)^2=4

Step-by-step explanation:

To find the curvature of the path we are going to use this formula:

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}

where

\boldsymbol{T}} is the unit tangent vector.

\frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| is the speed of the object

We need to find \boldsymbol{r}'(t), we know that \boldsymbol{r}(t)=cos \:2t \:\boldsymbol{i}+sin \:2t \:\boldsymbol{j}+ \:\boldsymbol{k} so

\boldsymbol{r}'(t)=\frac{d}{dt}\left(cos\left(2t\right)\right)\:\boldsymbol{i}+\frac{d}{dt}\left(sin\left(2t\right)\right)\:\boldsymbol{j}+\frac{d}{dt}\left(1)\right\:\boldsymbol{k}\\\boldsymbol{r}'(t)=-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}

Next , we find the magnitude of derivative of the position vector

|| \boldsymbol{r}'(t)}||=\sqrt{(-2\sin \left(2t\right))^2+(2\cos \left(2t\right))^2} \\|| \boldsymbol{r}'(t)}||=\sqrt{2^2\sin ^2\left(2t\right)+2^2\cos ^2\left(2t\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4\left(\sin ^2\left(2t\right)+\cos ^2\left(2t\right)\right)}\\|| \boldsymbol{r}'(t)}||=\sqrt{4}\sqrt{\sin ^2\left(2t\right)+\cos ^2\left(2t\right)}\\\\\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1\\\\|| \boldsymbol{r}'(t)}||=2\sqrt{1}=2

The unit tangent vector is defined by

\boldsymbol{T}}=\frac{\boldsymbol{r}'(t)}{||\boldsymbol{r}'(t)||}

\boldsymbol{T}}=\frac{-2\sin \left(2t\right)\boldsymbol{i}+2\cos \left(2t\right)\boldsymbol{j}}{2} =\sin \left(2t\right)+\cos \left(2t\right)

We need to find the derivative of unit tangent vector

\boldsymbol{T}'=\frac{d}{dt}(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j}) \\\boldsymbol{T}'=-2\cdot(\sin \left(2t\right)\boldsymbol{i}+\cos \left(2t\right)\boldsymbol{j})

And the magnitude of the derivative of unit tangent vector is

||\boldsymbol{T}'||=2\sqrt{\cos ^2\left(x\right)+\sin ^2\left(x\right)} =2

The curvature is

\kappa=\frac{||d\boldsymbol{T}/dt||}{ds/dt}=\frac{2}{2} =1

The tangential component of acceleration is given by the formula

a_{\boldsymbol{T}}=\frac{d^2s}{dt^2}

We know that \frac{ds}{dt}=|| \boldsymbol{r}'(t)}|| and ||\boldsymbol{r}'(t)}||=2

\frac{d}{dt}\left(2\right)\: = 0 so

a_{\boldsymbol{T}}=0

The normal component of acceleration is given by the formula

a_{\boldsymbol{N}}=\kappa (\frac{ds}{dt})^2

We know that \kappa=1 and \frac{ds}{dt}=2 so

a_{\boldsymbol{N}}=1 (2)^2=4

3 0
3 years ago
You and your four friends are going to a concert. The total price for the tickets was $118.75 which includes a $10 service fee.
Alex787 [66]

Answer:475

Step-by-step explanation:

118.75 is the cost whitch includes a 10$ service fee

118.75x10=475

8 0
3 years ago
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