<span><span>1/2<span>(<span>2g−3</span>)</span></span>=<span>−<span>4<span>(g+1)</span></span></span></span>
<span><span>1/2<span>(<span>2g−3</span>)</span></span>=<span>−<span>4<span>(g+1)</span></span></span></span>
<span><span>g+<span>−3/2</span></span>=<span><span>−4g</span>−4</span></span>
<span><span><span>g+<span>−3/2</span></span>+4g</span>=<span><span><span>−4g</span>−4</span>+4g</span></span><span><span>5g+<span>−32</span></span>=−4</span>
5g+−3/2+3/2=−4+3/2
<span><span>
5g</span>=<span>−5/2</span></span>
<span><span>5g/5</span>=<span><span>−52</span>5</span></span><span>
g=<span>−1<span>2
Hoped I helped!</span></span></span>
Step-by-step explanation:
that is only possible, if we don't need to use a 2 and a 5 (but especially the 2) to create the 1 in the middle (log(10) = log(2×5) or log (5×2) = 1).
in other words, if the 1 in the middle is just a given, and does not use any of our digits for is creation, then a solution is possible.
first of all, the upper left corner has to be log(0⁰). we cannot use the digit 0 for anything else. and 0^n (except for n = 0) is not a valid argument for a logarithm.
log(0⁰) = 0 log(2×1) = 0.3... log(9/4) = 0.35...
log(1×2) = 0.3... 1 log(3×6) = 1.26...
log(7/3) = 0.37... log(4×5) = 1.3... log(6⁹) = 7.0...
but as soon as we need to build the center 1 by log(10), which takes then away one of the 2s (as the only way to build 10 by multiplication is 2×5 or 5×2), there is no possible solution.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-10})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[x-(-6)]^2+[-4-(-10)]^2}\implies 10=\sqrt{(x+6)^2+(-4+10)^2} \\\\\\ 10^2=(x+6)^2+(6)^2\implies 100=x^2+12x+36+36 \\\\\\ 100=x^2+12x+72\implies 0=x^2+12x-28 \\\\\\ 0=(x+14)(x-2)\implies x= \begin{cases} -14\\ \boxed{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B-10%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7Bx%7D~%2C~%5Cstackrel%7By_2%7D%7B-4%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%2010%3D%5Csqrt%7B%5Bx-%28-6%29%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%2010%3D%5Csqrt%7B%28x%2B6%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%2010%5E2%3D%28x%2B6%29%5E2%2B%286%29%5E2%5Cimplies%20100%3Dx%5E2%2B12x%2B36%2B36%20%5C%5C%5C%5C%5C%5C%20100%3Dx%5E2%2B12x%2B72%5Cimplies%200%3Dx%5E2%2B12x-28%20%5C%5C%5C%5C%5C%5C%200%3D%28x%2B14%29%28x-2%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%20-14%5C%5C%20%5Cboxed%7B2%7D%20%5Cend%7Bcases%7D)
because B is on the IV Quadrant, the x-coordinate must be positive.
Answer:
It's the 3rd one using the formula 
Once you follow that step you need to multiply the exponent of 3 and 6 with the exponent outside the parenthesis..... which is why it's the 3rd one.
Don't be shy about using actual parentheses and commas.
Line through (1,8) perpendicular to
The perpendicular family is gotten by swapping the coefficients on x and y, negating exactly one of them. The constant is given directly by the point we're through:
Let's clear the fractions by multiplying both sides by 20.
Might as well stop here.
Answer: 25 x + 16 y = 153
