Answer:
y = - 
x + 2
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
y = 3x - 3 ← is in slope- intercept form
with slope m = 3
Given a line with slope m then the slope of a line perpendicular to it is
= - 
= - , hence
y = - x + c ← is the partial equation of the perpendicular line
To find c substitute (3, 1) into the partial equation
1 = - 1 + c ⇒ c = 1 + 1 = 2
y = - x + 2 ← equation of perpendicular line
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.
X=2 because first you have to combine like terms. Than you have to bring the Varible on one side. Finally you solve for the Varible and you get the answer2.
Answer:
the number
the type
Step-by-step explanation:
