The first one.
Hope this helps :)
Anonymous's answer is completely correct. I thought this problem was asking how to find the distance along the function from the point (2,2^8), and wrote the answer to that nice, tasty problem.
Simply integrate the line element with respect to some affine parameter!
<span><span>L=<span>∫10</span><span><span><span><span>(<span><span>∂x</span><span>∂λ</span></span>)</span>2</span>+<span><span>(<span><span>∂y</span><span>∂λ</span></span>)</span>2</span></span><span>−−−−−−−−−−−−−</span>√</span>dλ</span><span>L=<span>∫01</span><span><span><span>(<span><span>∂x</span><span>∂λ</span></span>)</span>2</span>+<span><span>(<span><span>∂y</span><span>∂λ</span></span>)</span>2</span></span>dλ</span></span>
In this case,
<span><span>x(λ)=λ(X−2)+2,</span><span>x(λ)=λ(X−2)+2,</span></span>
<span><span>y(λ)=(λ(X−2)+2<span>)8</span>.</span><span>y(λ)=(λ(X−2)+2<span>)8</span>.</span></span>
<span>Note that this approach can also solve the original problem, with some simplification.</span>
Answer:
Option C.
Step-by-step explanation:
Note: It is given question and option "=" sing is missing. On some places it should be "=" instead of "=".
Consider the given equation is
Using properties of \logarithm we get
![[\because \log_ax=b\Leftightarrow x=a^b]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Clog_ax%3Db%5CLeftightarrow%20x%3Da%5Eb%5D)
The solution is x=5.
To check the solution, substitute x=5 in the given equation.
Therefore, x-5 is a true solution because .
Answer:
Bottom right
Explanation:
visualize both triangle connected to the middle rectangle, and that's a triangular prism
