Do you mean 'arctan'?
arctan is same as 'atan' on this calculator.
If you want negative tangent of something, you just type in '-' and then 'tan'
Not enough information..... But I'd guess the sequence of Y is repeating 3, 6 , 3 , 6.......
So X=4 Y=6 X=5 Y=3 X=100 Y=6.
:)
Correct!!
My answer: \(28 = 1 + 9 + \sqrt9 \times 6\)
y = 0.3x - 7
Here :)
\(4 = x^{4/6}\\ x^{2/3} = 4\\ \sqrt[3]{x^2} = 4\\ x^2 = 64\\ x = \pm 8\)
4 (2x - 3)= 36
2x - 3 = 9
2x = 12
Try to solve it by yourself. It is just the last step remaining.
Use M.I.:
Let n = 1:
i^(2n + 1) = i^3 = -i........
The question is not correct...... Do you mean i^(4n + 1) = i?
I enjoyed Christmas because Mathematics is with me!! :)
Finally I have the answer!!!
\(x^{1/3}+(x-16)^{1/3}=(x-8)^{1/3}\\ x + x - 16 + 3 x^{1/3}(x-16)^{2/3}+3x^{2/3}(x-16)^{1/3}= x - 8\\ 3x^{1/3}(x-16)^{1/3}((x-16)^{1/3}+x^{1/3})=8-x\\ (x^{1/3}(x-16)^{1/3}(x-8)^{1/3})=\dfrac{8-x}{3}\\ x^3 - 24x^2 - 128x = \dfrac{(x-8)^3}{-27}\\ x^3 - 24x^2 - 128x = - \dfrac{x^3 + 24x^2 -64x + 512}{27}\\ 28x^3 - 672x^2 +(128\times 28-192)x - 512 = 0\\ 28x^3 - 672x^2 + 3648x - 512=0\\ (x-8)(7x^2 - 112x + 16)=0\\ x = 8\text{ or }8+12\sqrt{\dfrac{3}{7}}\text{ or }8-12\sqrt{\dfrac{3}{7}}\)
