now, the angle of θ, can only have a "y" value that is positive on, well, y is positive at 1st and 2nd quadrants
and "x" is positive only in 1st and 4th quadrants
now, that angle θ, can only have those two fellows, "y" and "x" to be positive, only in the 1st quadrant, and also both to be negative on the 3rd quadrant.
and that those two fellows, can also be both negative in the 3rd quadrant
3/1 = 3, and -3/-1 = 3
so, the solutions can only be "3", when both "y" and "x" are the same sign, and that only occurs on the 1st and 3rd quadrants
x/y + 3z^2 for x= 2/3, y= 6/7, and z=3
=2/3 / 6/7 + 3(3^2)
=2/3 * 7/6 + 3(9)
= 7/9 + 27
= 7/9 + 243/9
= 250/9
answer is a. 250/9
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.
