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3 years ago

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

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