Answer:
The intersection is
.
The Problem:
What is the intersection point of
and
?
Step-by-step explanation:
To find the intersection of
and
, we will need to find when they have a common point; when their
and
are the same.
Let's start with setting the
's equal to find those
's for which the
's are the same.

By power rule:

Since
implies
:

Squaring both sides to get rid of the fraction exponent:

This is a quadratic equation.
Subtract
on both sides:


Comparing this to
we see the following:



Let's plug them into the quadratic formula:




So we have the solutions to the quadratic equation are:
or
.
The second solution definitely gives at least one of the logarithm equation problems.
Example:
has problems when
and so the second solution is a problem.
So the
where the equations intersect is at
.
Let's find the
-coordinate.
You may use either equation.
I choose
.

The intersection is
.
13.45 will be the answer
since we are finding the hypotenuse, the formula is a^2 + b^2 = c^2
this will then be 10^2 + 9^2 = c^2
181=c^2
c=13.45 :)
Responder:
y = 200x + 2000
Explicación paso a paso:
Deje que la ecuación lineal se exprese como y = mx + c
m es la pendiente
c es la intersección
Si la empresa produce $ 10 unidades, sus costos totales son $ 4000 dólares y si produce $ 15 unidades sus costos ascienden a $ 5000 dólares, entonces podemos representar esto como puntos de coordenadas;
(10,4000) y (15,5000)
Conseguir la pendiente
m = 5000-4000 / 15-10
m = 1000/5
m = 200
Obtener la intersección c
Sustituya m = 2000 y el punto (10, 4000) en la fórmula y = MX + c
4000 = 200 (10) + c
4000-2000 = c
c = 2000
Sustituya m = 200 y c = 2000 en la ecuación y = mx + c
y = 200x + 2000
Por tanto, la expresión lineal requerida es y = 200x + 2000
Answer:
the decimal measuring system based on the meter, liter, and gram as units of length, capacity, and weight or mass. The system was first proposed by the French astronomer and mathematician Gabriel Mouton (1618–94) in 1670 and was standardized in France under the Republican government in the 1790s.
Step-by-step explanation:
it's 0.4 while it may look like 4% it's actually 40% of 1