Answer:
Step-by-step explanation:
Solve the inequality for v.3v-37>-2(8-5v)Simplify your answer as much as possible.
1 answer:
You might be interested in
I can't read that language but I'll guess it says write a second degree equation then solve for n when d=10.
Answer: n² - 3n - 20 = 0
That doesn't factor so there is no integer n solution.
That means there are no polygons with 10 diagonals.
The imaginary unit belongs to the set of complex numbers, denoted by
. These numbers take the form
, where
are any real numbers.
The set of real numbers, , is a subset of
, where each number in
can be obtained by taking
and letting
be any real number.
But any number in with non-zero imaginary part is not a real number. This includes
.
- "is it possible that i can use an imaginary number for a real number"
I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because .
There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising to the
-th power. Since
, we have