Let's say

a = 215,658

ok... so.. he first lost 1/3 of that

$\bf \textit{\underline{lost} }\frac{1}{3}a\qquad \qquad a-\cfrac{1}{3}a\implies a-\cfrac{a}{3}\implies \cfrac{3a-a}{3}\implies \cfrac{2a}{3}
\\\\\\
\textit{then he \underline{gained} }\frac{1}{2}\textit{ of }\frac{2a}{3}\qquad \qquad \cfrac{2a}{3}+\cfrac{\frac{2a}{3}}{2}\implies \cfrac{\frac{2a}{3}}{\frac{2}{1}}
\\\\\\
\cfrac{2a}{3}+\cfrac{2a}{3}\cdot \cfrac{1}{2}\implies \cfrac{2a}{3}+\cfrac{2a}{6}\implies \cfrac{4a+2a}{6}
\\\\\\
\cfrac{6a}{6}\implies \boxed{a}\implies 215,658$

so, his nickname might just be even steven, because, he ended up with the same original amount after the 1/2 bump.

4 0

2 years ago

Find three mutually orthogonal unit vectors in R^3 besides plusminus i, plusminus j, and plusminus k. There are multiple ways to

viktelen [127]

Answer:

u=(1,1,2), v=(-3,1,2), w=(1,-2,1/2)

Step-by-step explanation:

Since the vectors must be orthogonal, then they must satisfy that the dot product between them is 0.

Then

$1(x)+1(-1)+2(2)=0\\x=-3$

and $-3(1)-1y+2z=0 \text{ and }\\1+1y+2z=0\\$ .

Solving for y, y=-2z-1 and substituting in the first equation, -3+2z+1+2z=0, then

z=1/2, and y=-2(1/2)-1=-2

6 0

2 years ago