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marta [7]
3 years ago
7

Please answer Find the area of the polygon. 6 cm6 cm10 cm

Mathematics
1 answer:
Monica [59]3 years ago
4 0

Answer:

Area: 360 cm

Perimeter: 44 cm

Step-by-step explanation:

To calculate the area, you must multiply the length x width.

If you need help calculating the perimeter too, just add all your lengths and widths together twice.

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Find the value of x .
disa [49]

Answer:

-3

Step-by-step explanation:

3 0
3 years ago
Renae has a checking account balance of -$40.75. Paula has a checking account balance of -$25.20. Who owes more?
Volgvan

Answer:

Renae owes 15.55 more.


Step-by-step explanation: 40.75 - 25.20 = 15.55



7 0
3 years ago
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A choreographer uses a number line to position dancers for a ballet. Dancers A and B
kenny6666 [7]
First we will calculate the distance AB. 
AB=B-A=23-5=18.
If C is 5/9 of the distance, mathematically speaking that means:
C= \frac{5}{6} *AB= \frac{5}{6} *18=15.
R: The coordonate of C is 15.
5 0
3 years ago
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The Tetrahedron Superyacht designed by Jonathan Schwinge is a regular tetrahedron with edges 20 m long. The distance r from one
shusha [124]
A tetrahedron is a triangular pyramid. It has four triangular faces and four vertices. 

In the question, we are told that the length of the edges are all equal = 20m and the length from one corner to the center of the base = 11.5m

We sketch a diagram of a tetrahedron with the given measurement as shown below

To find the height, we will use the Pythagoras theorem

20² = h² + 11.5²
h² = 20² - 11.5²
h² = 267.75
h = √267.75
h = 16.36 (rounded to two decimal places)

8 0
3 years ago
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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
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