Difference between revisions of "User:Jiseop55406"

(Replaced content with "amogus")
(Tag: Replaced)
(Undo revision 176249 by Jiseop55406 (talk))
(Tag: Undo)
Line 1: Line 1:
amogus
+
Problem
 +
The ratio <math>\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}</math> is closest to which of the following numbers?
 +
 
 +
<math>\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 10</math>
 +
 
 +
== soulition 1 ==
 +
Helo,
 +
<cmath>
 +
\begin{align*}
 +
\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}} &= \frac{10^{2000}(1+10^2)}{10^{2000}(10+10)}\\
 +
&= \frac{101}{20}\\
 +
&= 5.05,
 +
\end{align*}
 +
</cmath>
 +
soo basicly the ansiwer is obviously <math>\text{(F)} \ 420.69</math>. <math>\mathbf{Q.E.D}</math>.<math>\blacksquare</math>
 +
 
 +
~Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society

Revision as of 22:29, 4 August 2022

Problem The ratio $\frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}}$ is closest to which of the following numbers?

$\text{(A)}\ 0.1 \qquad \text{(B)}\ 0.2 \qquad \text{(C)}\ 1 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 10$

soulition 1

Helo, \begin{align*} \frac{10^{2000}+10^{2002}}{10^{2001}+10^{2001}} &= \frac{10^{2000}(1+10^2)}{10^{2000}(10+10)}\\ &= \frac{101}{20}\\ &= 5.05, \end{align*} soo basicly the ansiwer is obviously $\text{(F)} \ 420.69$. $\mathbf{Q.E.D}$.$\blacksquare$

~Foogle and Hoogle, Members of the Ooga Booga Tribe of The Caveman Society