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White raven [17]
3 years ago
8

I’ll give u brainliest!! what is the area of shape 1 2 and 3

Mathematics
2 answers:
Fed [463]3 years ago
6 0

Answer:

yes

Step-by-step explanation:

DedPeter [7]3 years ago
5 0

Answer:

Shape 1: 197.04

Shape 2: 125.44

Shape 3: 62.72

Step-by-step explanation:

For shape 1, you first need to find the radius to then find the area of a circle. Which is 394.08. Since this isn't a full circle, you will need to divide it by 2 to get 197.04. For shape 2(Easiest to find) You do 11.2*11.2 to get the answer of 125.44. And lastly for shape 3, you can basically do the same as what you did for shape 2. But since it is a triangle, you divide your answer by 2, to get the answer of 62.72.

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Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




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