Difference between revisions of "User:Jiseop55406"

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

Revision as of 20:57, 29 June 2022

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

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

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 $\text{(F)} \ 420.69$. $\mathbf{Q.E.D}$.$\blacksquare$