Use the formula x+x+1+x+2 or 3x+3=170 to find x. x is the first number. Then add x+1 to find the second number. Then add x+2 to find the third final number.
x will equal 167/3 which is a decimal, so just plug it in to the 3 equations I mentioned above to find your 3 consecutive numbers. More questions? Just ask me!
Answer:
Tray will overflow with
of paint.
Step-by-step explanation:
Dimensions the tray is 10 inch by 14 inch by 4 cm


Volume the tray can hold is


The volume of paint Billy has is 
Difference in the volume of paint and volume of tray in cubic inches is

The tray will overflow with
of paint.
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
.
Y^3 * y^5 = y^8. you add the exponents