We are talking real numbers here so the number under the square root must be positive.
$$\\2sinx-\sqrt3\ge0\\\\ 2sinx\ge\sqrt3\\\\ sinx\ge\frac{\sqrt3}{2}\\\\$$
for starters sinx must be positive so it must be 1st and 2nd quad only.
$$asin(\frac{\sqrt3}{2})=\frac{\pi}{3}$$ the second quadrant equivalent is $$\pi-\frac{\pi}{3}=\frac{2\pi}{3}$$
$$\frac{\pi}{3}\le x \le \frac{2\pi}{3}$$
I know this because sin(pi/2)=1 and I know what the sine curve looks like so I know the answer will be between these two values.
This answer is good.
Well, you would show the working the same as I did.
But the question actually asks you for where it is continuous.
So the answer would be
continuous when
x<-4, -4<x<0, 0<x<4, and x>4
OR
$$(-\infty,-4), (-4,0), (0,4), (4,+\infty)$$
I am not mad either but please Dragon no more questions for a while.
Dragon, please only comment when you can be helpful. You know chilledz was not talking to you and that time you had nothing helpful to say!
Your gradient post was okay, Chilledz might have found that helpful.
Yes, it is the slope OF the tangent line at x=5.
Not at the moment dragon, I have other things to do. Sorry
I have already answered about 100 questions for you this morning dragon.
$$\\2n^4\times 5n^4\\\\ =2*n*n*n*n*5*n*n*n*n\\\\ =2*5*n*n*n*n*n*n*n*n\\\\ =10n^8$$
$$\\f(x)= \frac{\sqrt{x+10}}{x^{3}-16x}\\\\ f(x)= \frac{\sqrt{x+10}}{x(x^{2}-16)}\\\\ f(x)= \frac{\sqrt{x+10}}{x(x-4)(x+4)}\\\\ $You cannot divide by 0 so x cannot be 0,4 or -4$\\\\ $The function is discontinuous at x=0, x=4 and x=-4$\\\\ $In latex, the backslash indicates that a function will follow, $\\ $the arguments of the function usually have to go in parentheses. $\\ $The only exception is some very simple functions where it is assumed$\\ $that just the next entry is the argument. $ ex. \sqrt9$$
that is correct - but you can leave out the times signs.
3t-x+8y-t+2
you cannot add the 2 to the letters - just leave that number alone.
12yaks and 5 xylophones
a xylophone is the same as 1 xylophone.
Here is one
6y+9t-y+2y+z (t can be trains if you like)