Asked by Sannu Xeno
1.solve the equations below
x+y+z=1
x^2+y^2+z^2=35
x^3+y^3+z^3=97
2.Determine the area of the largest rectangle that can be inscribed in the circle x^2+y^2=a^2.
Thus, name the rectangle formed
3.a ship leaves airport and travels 25kmnon a bearing of 037° and then 60km on a bearing of 307°.calculate the distance from the port
4.The last digit of 2016^2017 is••••••••
5. Simplify 3√4^13?
x+y+z=1
x^2+y^2+z^2=35
x^3+y^3+z^3=97
2.Determine the area of the largest rectangle that can be inscribed in the circle x^2+y^2=a^2.
Thus, name the rectangle formed
3.a ship leaves airport and travels 25kmnon a bearing of 037° and then 60km on a bearing of 307°.calculate the distance from the port
4.The last digit of 2016^2017 is••••••••
5. Simplify 3√4^13?
Answers
Answered by
Reiny
One of the ways is to square and cube the first equation then substituting, but it gets pretty scary.
Lets use the "Just Look At It Theorem".
Look at x^2 + y^2 + z^2 = 35, assuming that there is an integer solution as there usually is in this type of question, all of the variables must be between -5 and +5
suppose x = 5, then 25 + y^2 + z^2 = 35, or y^2 + z^2 = 10
It should be easy to see that the only two squares that add up to 10 are 3^2 ans 1^2, thus y could be 3 and z = 1
So (5,3,1) would be a solution to the 2nd equation, (but so would (3,5,1) etc )
Can we make these work in the 1st ?
Mmmmhhhh, after about 20 seconds of consideration, it must be (5, -3, -1)
Well, look at this, that triple also works in the 3rd eqution:
5^3 + (-3)^3 + (-1)^3
= 125 - 27 - 1 = 97
Because of the symmetry of the equations, the 5,-3,-1 can be arranged in 3! ways, to have 6 possible solutions.
e.g. (-1,5,-3) would also work
Lets use the "Just Look At It Theorem".
Look at x^2 + y^2 + z^2 = 35, assuming that there is an integer solution as there usually is in this type of question, all of the variables must be between -5 and +5
suppose x = 5, then 25 + y^2 + z^2 = 35, or y^2 + z^2 = 10
It should be easy to see that the only two squares that add up to 10 are 3^2 ans 1^2, thus y could be 3 and z = 1
So (5,3,1) would be a solution to the 2nd equation, (but so would (3,5,1) etc )
Can we make these work in the 1st ?
Mmmmhhhh, after about 20 seconds of consideration, it must be (5, -3, -1)
Well, look at this, that triple also works in the 3rd eqution:
5^3 + (-3)^3 + (-1)^3
= 125 - 27 - 1 = 97
Because of the symmetry of the equations, the 5,-3,-1 can be arranged in 3! ways, to have 6 possible solutions.
e.g. (-1,5,-3) would also work