What are some easy ways to find the value of (2017^4−2016^4)/(2017^2+2016^2) without calculator
1 answer:
Answer:
4033
Step-by-step explanation:
An easy way to solve this problem is to notice the numerator, 2017^4-2016^4 resembles the special product a^2 - b^2. In this case, 2017^4 is a^2 and 2016^4 is b^2. We can set up equations to solve for a and b:
a^2 = 2017^4
a = 2017^2
b^2 = 2016^4
b = 2016^2
Now, the special product a^2 - b^2 factors to (a + b)(a - b), so we can substitute that for the numerator:
<h3>We can notice that both the numerator and denominator contain 2017^2 + 2016^2, so we can divide by which is just one, and will simplify the fraction to just:
2017^2 - 2016^2
This again is just the special product a^2 - b^2, but in this case a is 2017 and b is 2016. Using this we can factor it:
(2017 + 2016)(2017 - 2016)
And, without using a calculator, this is easy to simplify:
(4033)(1)
4033
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Answer:
0 ≤ Ф ≤ 4π.
Step-by-step explanation:
since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))
the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.
We take
Note that
Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.
The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.