Without knowing what the functions are, you have to leave it as a sum. For example, f(x)+g(x) is as far as you can get, if you don't know what f(x) or g(x) are.
If you are told f(x)=x+5, and g(x)=2x-1, you can plug in to get x+5+2x-1, which just equals 3x+4.
To calculate two-thirds of a sum, you first find the sum of the numbers. Then, you multiply the sum by 2/3, which is the same as multiplying the sum by 2 and dividing by 3. This will give you the result of two-thirds of the sum. For example, if the sum is 15, two-thirds of the sum would be (15 x 2) / 3 = 10.
How about: 18 and 24
224, 225
2 and 15 5 and 12
Three and Seven
yes
You find the sum of two numbers by adding the two together. Example: The sum of 1 + 1 is 2
Yes.
A generalization of wavelets that uses vector valued functions as scaling and wavelet functions. The two scale equation contains a sum of matrices instead of sum of scalars.
To find the median with two numbers, do as you would to find the "average" or the "mean" by finding the sum between the two numbers and dividing the sum (or total) by two.
Well, it sounds like a plausible statement, and maybe it would be true . But we haveno idea what the graph of two functions is.Perhaps you could graph the sum of two functions, or the difference of two functions,or their product, or their quotient. We believe that if the original two functions areboth continuous, then their sum and difference would also be continuous, but theirproduct and their quotient might not necessarily be continuous. However, we stilldon't know what the "graph of two functions" is.
Combining two or more numbers to find the sum is called addition.
d/dx (u + v) = du/dx + dv/dx essentially this means that if you are finding the derivative of two functions u and v that you can find the derivative of each function separately and then add the derived functions to get the answer.
in a positive
Yes. You would have to multiply to change it.
Find the two numbers with the largest magnitudes (absolute values). The sum of their squares will be the maximum.
1