\(3x+2y=17 \\ 2x -2y=-2 \\ \mbox{add the two equations to obtain}\\ 5x=15\\ x=3, \mbox{ and from the first equation, } y=4\)
The human body isn't one homogeneous substance.
Fat probably melts around 100degF.
Bones don't melt at all.
Actually Melody you are correct. I'm mistaken.
From Wiki
"In a number with a decimal point, trailing zeros, those to the right of the first non-zero digit, are significant."
so 5 is the correct answer.
\(x^2 + 6x + 17 =0 \\ x^2 + 6x +9 -9 + 17 = 0\\ (x+3)^2+8=0 \\ (x+3)^2 = -8 \\ \mbox{and this clearly has no solution over the real numbers as any number squared is non-negative.}\)
\(\mbox{If we solve this over the complex numbers we get}\\ (x+3)^2=-8 = (-1)8 \\ (x+3) = \pm \sqrt{-1}\sqrt{8} = \pm \imath \sqrt{8} = \pm \imath 2\sqrt{2} \\ x = -3 \pm \imath 2\sqrt{2}\)
The 3 and the 1 are the significant digits here so just 2.
\((y+2)^2 = 10^2 + y^2 \\ y^2 + 4y + 4 = 100+y^2\\ 4y+4=100 \\ 4y=96 \\ y=24\)
\(\sqrt{27 v^{10}}=\sqrt{3(3v^5)^2} = 3v^5\sqrt{3}\)
\(5^{2x+3}=6^{x+1} \\ (2x+3)\log(5)=(x+1)\log(6) \\ (2\log(5)-\log(6))x=\log(6)-3\log(5) \\ x = \dfrac{\log(6)-3\log(5)}{2\log(5)-\log(6)}\)
\(\mbox{This is a Poisson distribution problem.}\\ \mbox{With 0.7 flaws per 100 feet we will have 7 flaws per 1000 feet.}\)
\(\mbox{So the parameter }\lambda = 7\\ \mbox{We evaluate the Poisson distribution PDF at 3} \\ P[\mbox{3 flaws in 1000 feet}]=\dfrac {7^3} {3!}e^{-7} = 0.05213\)
\(-\dfrac 7 x + 5 = -3 \\ -\dfrac 7 x = -8 \\ -7 = -8x \\ x = \dfrac 7 8\)
+6251 