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Hello,
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
![l= 2*5* \int\limits^{\pi}_0{sin( \frac{t}{2} )} \, dt= 5[-cos (t/2)]_0^{\pi}\\\\ =5(0+1)=5](https://tex.z-dn.net/?f=l%3D%202%2A5%2A%20%5Cint%5Climits%5E%7B%5Cpi%7D_0%7Bsin%28%20%5Cfrac%7Bt%7D%7B2%7D%20%29%7D%20%5C%2C%20dt%3D%205%5B-cos%20%28t%2F2%29%5D_0%5E%7B%5Cpi%7D%5C%5C%5C%5C%0A%3D5%280%2B1%29%3D5)
Here is the method but i may have make some mistakes.
r=5(1+cos t)
r'=5(-sin t)
r²+r'²= 25[(1+cos t)²+(-sin t)²]=50(1-cos t)=50 sin² (t/2)
Between 0 and π, sin x>0 ==>|sin x|=sin x
Here is the method but i may have make some mistakes.
The equation that has an infinite number of solutions is
An equation that has an infinite number of solutions would be in the form
a = a
This means that both sides of the equation would be the same
Start by simplifying the options
3(x – 1) = x + 2(x + 1) + 1
3x - 3 = x + 3x + 2 + 1
3x - 3 = 4x + 3
Evaluate
x = 6 ----- one solution
x – 4(x + 1) = –3(x + 1) + 1
x - 4x - 4 = -3x - 3 + 1
-3x - 4 = -3x - 2
-4 = -2 ---- no solution
2x + 3 = 2x + 1 + 2
2x + 3 = 2x + 3
Subtract 2x
3 = 3 ---- infinite solution
Hence, the equation that has an infinite number of solutions is
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<u>Complete question</u>
Which equation has infinite solutions?
3(x – 1) = x + 2(x + 1) + 1
x – 4(x + 1) = –3(x + 1) + 1