Solve for x. Round your answer to the nearest tenth. Will brainlist first correct
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Domain of a function
<h2>STEP 3: finding the x values that make x²-6x-55>0 (factoring)</h2>
<h2>STEP 4: finding the x values that make (x+5)(x-11)>0 (factoring)</h2>
We want to find the domain of the following function:
This means that we want to find the x-values that it can take.
<h2>STEP 1: analyzing the simplies form of the function</h2>Let's analyze the simpliest form of the function:
Its graph is:
Then, for the simpliest form of the function, the x-values can only be higher than 0.
This means that its domain is
domain = x > 0
<h2>STEP 2: domain of the given function</h2>Based on the above we can deduce that for the <em>ln(x)</em> function, what is inside the parenthesis should be higher than 0 on this kind of functions.
This is that for
then
In order to find the values of x that make
we must factor it.
We want to find a pair of numbers that when multiplied give the last term (-55) and when added together give the second term (-6).
For the last term of the polynomial: -55, we have that
(-5) · 11 = 55
5 · (-11) = 11
If we add them:
-5 + 11 = 6
5 - 11 = -6
The pair of numbers that when multiplied give the last term (-55) and when added together give the second term (-6), are: 5 and -11
We use them to factor the polynomial:
Then,
In order to find them, we are going to separate the factors (x+5) and (x-11) and analyze when they are positive or negative:
Combining them:
Since we are going to multiply both factors:
We use the diagram to analyze the sign of their product:
Then
when x < -5 and when x > 11. This is the domain.
Therefore, expressed in set notation:
domain = {x|x∈(-∞, -5)∪(11, ∞)}
<h2>Answer: domain = {x | x ∈ (-∞, -5)∪(11, ∞)}</h2>
To solve this problem, set up and solve a system of equations. The variables b and m will represent a bread loaf and milk jug, respectively:
I would solve using substitution. Take one of the equations and set it equal to one of the variables, for example:
Now, plug this into the other equation for m and solve for b:
We now know that a loaf of bread costs $2.50. Plug this value in for b in the first equation and solve for m:
One jug of milk costs $1.50 and one loaf of bread costs $2.50.
I would solve using substitution. Take one of the equations and set it equal to one of the variables, for example:
Now, plug this into the other equation for m and solve for b:
We now know that a loaf of bread costs $2.50. Plug this value in for b in the first equation and solve for m:
One jug of milk costs $1.50 and one loaf of bread costs $2.50.