Answer:
Increasing on it's domain 
because the slope is positive.
The domain and range are both all real numbers, also known as
.
Step-by-step explanation:
All domain really means is what numbers can you plug in and you get number back from your function.
I should be able to plug in any number into 3x+2 and result in a number. There are no restrictions for x on 3x+2.
The domain is all real numbers.
In interval notation that is .
Now the range is the set of numbers that get hit by y=3x+2.
Well y=3x+2 is a linear function that is increasing. I know it is increasing because the slope is positive 3. I wrote out the positive part because that is the item you focus on in a linear equation to determine if is increasing or decreasing.
If slope is positive, then the line is increasing.
If slope is negative, then the line is decreasing.
So y=3x+2 hits all values of y because it is increasing forever. The range is all real numbers. In interval notation that is .
The answer is A
1/4, 1/4, 1/4, 1,4...
Answer:
m - 10
Step-by-step explanation:
7 plus = 7 +
The sum of m and -17 = m + (-17)
7 + (m + -17)
We can get rid of the parentheses
7 + m + -17
Combine like terms
7 + -17 = -10
m + -10
Get rid of the addition sign
m - 10
Answer: The required solution of the given IVP is
Step-by-step explanation: We are given to find the solution of the following initial value problem :
Let 
be an auxiliary solution of the given differential equation.
Then, we have
Substituting these values in the given differential equation, we have
![m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.](https://tex.z-dn.net/?f=m%5E2e%5E%7Bmx%7D-e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20%28m%5E2-1%29e%5E%7Bmx%7D%3D0%5C%5C%5C%5C%5CRightarrow%20m%5E2-1%3D0~~~~~~~~~~~~~~~~~~~~~~~~~~%5B%5Ctextup%7Bsince%20%7De%5E%7Bmx%7D%5Cneq0%5D%5C%5C%5C%5C%5CRightarrow%20m%5E2%3D1%5C%5C%5C%5C%5CRightarrow%20m%3D%5Cpm1.)
So, the general solution of the given equation is
where A and B are constants.
This gives, after differentiating with respect to x that
The given conditions implies that
and
Adding equations (i) and (ii), we get
From equation (i), we get
Substituting the values of A and B in the general solution, we get
Thus, the required solution of the given IVP is
Answer:
an equation that involves some ordinary derivatives (as opposed to partial derivatives
