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Lyrx [107]
3 years ago
15

If you have 14 feet of fencing, what are the areas of the different rectangles you could enclose with the fencing? Consider only

whole number dimensions
Mathematics
1 answer:
disa [49]3 years ago
8 0

Answer:

Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

Step-by-step explanation:

Feets of fencing = 14

Perimeter = 14

Let x = Length y = width

Perimeter = 2(x + y)

For x = 1

2(1 + y) = 14

1 + y = 7

y =6

Area of rectangle ; x *y = 1 * 6 = 6 ft²

For x = 2

2(2 + y) = 14

2 + y = 7

y = 5

Area of rectangle ; x *y = 2 * 5 = 10 ft²

For x = 3

2(3 + y) = 14

3 + y = 7

y =4

Area of rectangle ; x *y = 3 * 4 = 12 ft²

For x = 4

2(1 + y) = 14

4 + y = 7

y = 3

Area of rectangle ; x *y = 4 * 3 =12 ft²

Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

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You stand a known distance from the base of the tree, measure the angle of elevation the top of the tree to be 15â—¦ , and then
gogolik [260]

Answer:

The maximum possible error of in measurement of the angle is  d\theta_1  =(14.36p)^o

Step-by-step explanation:

From the question we are told that

    The angle of elevation  is  \theta_1  =  15 ^o =  \frac{\pi}{12}

     The height of the tree is  h

      The distance from the base is  D

h is mathematically represented as

            h  = D tan \theta       Note : this evaluated using SOHCAHTOA i,e

                                               tan\theta  =  \frac{h}{D}

Generally for small angles the series approximation of  tan \theta \  is

          tan \theta  =  \theta  + \frac{\theta ^3 }{3}

So given that \theta =  15 \ which \ is \ small

       h = D (\theta + \frac{\theta^3}{3} )

       dh = D (1 + \theta^2) d\theta

=>        \frac{dh}{h} =  \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta

Now from the question the relative error of height should be at  most

        \pm  p%

=>    \frac{dh}{h} =   \pm p

=>    \frac{1 + \theta ^2}{\theta + \frac{\theta^3}{3} } d \theta  = \pm p

=>      d\theta  =  \pm  \frac{\theta +  \frac{\theta^3}{3} }{1+ \theta ^2} *    \ p

 So  for   \theta_1

            d\theta_1  =  \pm  \frac{\theta_1 +  \frac{\theta^3_1 }{3} }{1+ \theta_1 ^2} *    \ p

substituting values  

          d [\frac{\pi}{12} ]  =  \pm  \frac{[\frac{\pi}{12} ] +  \frac{[\frac{\pi}{12} ]^3 }{3} }{1+ [\frac{\pi}{12} ] ^2} *    \ p

 =>       d\theta_1  = 0.25 p

Converting to degree

           d\theta_1  = (0.25* 57.29) p

            d\theta_1  =(14.36p)^o

4 0
3 years ago
What is the area of a kite in the figure below?
Leto [7]

Answer:

The are is 1,200 cm squared

Step-by-step explanation:

10 * 20 = 200

20 * 50 = 1000

1000 + 200 = <em><u>1,200cm squared</u></em>

<em><u></u></em>

<em><u>Hope this helped! Have a nice day! Plz mark as brainliest!!!</u></em>

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5 0
3 years ago
2. If y(x-1)=z then x=
Mashcka [7]
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x=(z+y)/y
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timofeeve [1]

Answer:

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zhenek [66]
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