Answer:
Phitagors theory
Step-by-step explanation:
a^2=b^2=c^2
I think it would be D, about 200, because 243 is closer to 200 then it is any of the other choices
Answer:
use logarithms
Step-by-step explanation:
Taking the logarithm of an expression with a variable in the exponent makes the exponent become a coefficient of the logarithm of the base.
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You will note that this approach works well enough for ...
a^(x+3) = b^(x-6) . . . . . . . . . . . variables in the exponents
(x+3)log(a) = (x-6)log(b) . . . . . a linear equation after taking logs
but doesn't do anything to help you solve ...
x +3 = b^(x -6)
There is no algebraic way to solve equations that are a mix of polynomial and exponential functions.
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Some functions have been defined to help in certain situations. For example, the "product log" function (or its inverse) can be used to solve a certain class of equations with variables in the exponent. However, these functions and their use are not normally studied in algebra courses.
In any event, I find a graphing calculator to be an extremely useful tool for solving exponential equations.
Point c because its the top and they rotated it to it's side
Solution in the attachment.
![\det\left[\begin{array}{ccc}4&2&-3\\5&-6&1\\0&2&-1\end{array}\right]=4\cdot\left|\begin{array}{ccc}-6&1\\2&-1\end{array}\right|-2\cdot\left|\begin{array}{ccc}5&1\\0&-1\end{array}\right|-3\cdot\left|\begin{array}{ccc}5&-6\\0&2\end{array}\right|](https://tex.z-dn.net/?f=%20%5Cdet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%26-3%5C%5C5%26-6%261%5C%5C0%262%26-1%5Cend%7Barray%7D%5Cright%5D%3D4%5Ccdot%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D-6%261%5C%5C2%26-1%5Cend%7Barray%7D%5Cright%7C-2%5Ccdot%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D5%261%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%7C-3%5Ccdot%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D5%26-6%5C%5C0%262%5Cend%7Barray%7D%5Cright%7C%20)
