The length of the function y = 3x over the given interval [0, 2] is 3.2 units
For given question,
We have been given a function y = 3x
We need to find the length of the function on the interval x = 0 to x = 2.
Let f(x) = 3x where f(x) = y
We have f'(x) = 3, so [f'(x)]² = 9.
Then the arc length is given by,
![\int\limits^a_b {\sqrt{1+[f'(x)]^2} } \, dx\\\\= \int\limits^2_0 {\sqrt{1+9} }\, dx\\\\=\sqrt{10}\\\\ =3.2](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Csqrt%7B1%2B%5Bf%27%28x%29%5D%5E2%7D%20%7D%20%5C%2C%20dx%5C%5C%5C%5C%3D%20%5Cint%5Climits%5E2_0%20%7B%5Csqrt%7B1%2B9%7D%20%7D%5C%2C%20dx%5C%5C%5C%5C%3D%5Csqrt%7B10%7D%5C%5C%5C%5C%20%3D3.2)
This means, the arc length is 3.2 units.
Therefore, the length of the function y = 3x over the given interval [0, 2] is 3.2 units
Learn more about the arc length here:
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Answer:
x^2 y^2
Step-by-step explanation:
9x^2y^2/x^2y^2=9
5x^2y^2/x^2y^2=5
since 9 and 5 have no common factors except 1, x^2 y^2
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Answer:
22,500 =2p
Step-by-step explanation:
Because the price of the new car is the price of the old car times 2.
22,500 =2p
First you get P alone. P is being multiplied by 2 so you divide it by 2 so it cancels out.
11250 = p
When you do something to one side of the equation you do the same to the other side so they are still equal to each other.
