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aksik [14]
3 years ago
10

A lot for sale is shaped like a trapezoid. The bases of the trapezoid represent the widths of the front and back yards. The widt

h of the back yard is twice the width of the front yard. The distance from the front yard to the back yard, or the height of the trapezoid, is equal to the width of the back yard. Find the width of the front and back yards, given that the area is 3,300 square feet. Round to the nearest foot
Mathematics
1 answer:
RoseWind [281]3 years ago
4 0

Answer:

Front: 33 ft

Back: 66 ft

Step-by-step explanation:

Let w be the width of the front yard

½[w + 2w]×2w = 3300

3w² = 3300

w² = 1100

w = 33.1662479036

w = 33 ft

2w = 66 ft

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Paha777 [63]

The product of the given expression is -120. Option A is correct

<h3>Product of negative numbers.</h3>

Given the expression below as shown;

−5⋅(−3)⋅(−8)

Since the product of two negative number is equal to positive, then;

−5⋅(−3)⋅(−8) = -8(5*3)

Simplify the result to have:

-8 * 15

find the final product

-8 * 15 = -8(5+10)

-8 * 15 = -40 - 80

-8 * 15 = -120

Hence the product of the given expression is -120

Learn more on product of negative numbers here: brainly.com/question/17485302

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1 year ago
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}

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|| \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

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\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

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The curvature is

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

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a_{\boldsymbol{T}}=0

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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

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