Assume that you only include whole numbers (1,2,3,4,5,6,7,8,9) and not 3.5 and such
so if 1 is odd and less than 5 then it is
1 or 3, since 5 isn't included
then the other number, to be less than 5 when added,
must be
1+x<5
3+x<5
solve each
1+x<5
subtract 1
x<4
set of answers are 1,2,3
3+x<5
subtract 3
x<2
set of answer is 1
so the possible numbers are
1,2,3
that is 3 numbesr out of 9 so
probability=(total desired outcomes)/(total possible outcomes) so
disred outcomes=3
total possible=9
3/9=1/3
the probabiltiy is 1/3
This is a strange question, and f(x) may not even exist. Why do I say that? Well..
[1] We know that f(a+b) = f(a) + f(b). Therefore, f(0+0) = f(0) + f(0). In other words, f(0) = f(0) + f(0). Subtracting, we see, f(0) - f(0) = f(0) or 0 = f(0).
[2] So, what's the problem? We found the answer, f(0) = 0, right? Maybe, but the second rule says that f(x) is always positive. However, f(0) = 0 is not positive!
Since there is a contradiction, we must either conclude that the single value f(0) does not exist, or that the entire function f(x) does not exist.
To fix this, we could instead say that "f(x) is always nonnegative" and then we would be safe.
Answer:
Step-by-step explanation:
<u>Given Second-Order Homogenous Differential Equation</u>
<u>Use Auxiliary Equation</u>
<u />
<u>General Solution</u>
<u />
Note that the DE has two distinct complex solutions where and are arbitrary constants.
Answer:
x = -10
Step-by-step explanation:
y = 2x + 5
-15 = 2x + 5
-20 = 2x
-10 = x