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you can't always jump to conclusions like that just by looking at the graph. :)

anyway

the distance from the graph to the origin can be represented like this (by the Pythagorean theorem):

$\sqrt{x^2+(\frac{1}{\sqrt{2}}(x^2-3))^2}\\=\sqrt{x^2+\frac{(x^2-3)^2}{2}}\\= \sqrt{x^2+\frac{x^4-6x^2+9}{2}}\\=\sqrt{\frac{x^4-4x^2+9}{2}}\\ =\sqrt{\frac{x^4-4x^2+4+9-4}{2}}\\=\sqrt{\frac{(x^2-2)^2+5}{2}}$

Since $(x^2-2)^2$ will always be nonnegative for real numbers, the best you can do to minimize it is to set it equal to 0, and you will obtain your final answer:

$\sqrt{\frac{0+5}{2}}=\boxed{\sqrt{\frac{5}{2}}}$

for completeness, if you want to know the x value for which this minimum distance occurs, you can just solve an equation:

$(x^2-2)^2=0\\x^2-2=0\\x=\pm\sqrt{2}$

lol sorry i got it wrong, just noticed it

OP plz change the best answer

My answer was supposed to be:

$12^2+x^2=(12+16)^2\\x^2=640\\x=\sqrt{640}=8\sqrt{10}$

and then multiply by 2

$\frac{2x^2 + 3x - 5}{x(x^2 - 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}$

multiply both sides by $x(x^2-1)$:

$2x^2+3x-5=A(x-1)(x+1)+B(x)(x+1)+C(x)(x-1)$

Notice that if you set x to equal 0, 1, and -1, respectively, you can cancel out two of the 3 coefficients, which will make it easier to solve. First, set x = 0:

$-5=-A\\A=5$

Now set x = 1:

$2+3-5=2B\\B=0$

Lastly, set x = -1:

$2-3-5=2C\\ -6=2C\\C=-3$

Therefore, the answer is $\boxed{(5, 0, -3)}$

$\frac{1}{x} + \frac{1}{y} = \frac{1}{18}\\ \frac{x+y}{xy} = \frac{1}{18}\\ xy=18x+18y\\ xy-18x-18y=0\\ (x-18)(y-18)=324$

The pair of numbers that multiplies to 324 that are closest to each other but are not equal to each other are 12 and 27, so the answer is (18+12)+(27+18)

btw a very similar question was already answered here:

You could also look up "method of differences" or "Lagrange interpolation" for faster computation. I'm not really familiar with it, but they both look faster than solving 6 simultaneous equations.

Since we need to compute the last 3 digits, we can take the whole expression mod 1000:

$1991\cdot1993\cdot1995\cdot1997\cdot1999\cdot2001\cdot2003 (\text{mod 1000})\\ =-9\cdot-7\cdot-5\cdot-3\cdot-1\cdot1\cdot3 (\text{mod 1000})\\ =-2835 (\text{mod 1000})\\ =165(\text{mod 1000})$

The sum of the hundreds, tens, and units digit is $1+6+5=\boxed{12}$

Notice that $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$. You could multiply the second equation by 2 and then add the equations up:

$x^2+y^2+z^2+2xy+2yz+2zx=9+6\cdot2\\(x+y+z)^2=21$

Take the square root of both sides and you will arrive at the answer.

(notice that the negative square root is extraneous since x, y, and z are all nonnegative)

1356 leaves a remainder of 6 when divided by 27 (you can check using long division.

The problem reduces to: $3n \equiv 6 (\text{mod 27})$. In other words, what is the smallest number that when multiplied by 3 is equal to 6?