The answer is 26mm, hope this helps!
Given that JKLM is a rhombus and the length of diagonal KM=10 na d JL=24, the perimeter will be found as follows;
the length of one side of the rhombus will be given by Pythagorean theorem, the reason being at the point the diagonals intersect, they form a perpendicular angles;
thus
c^2=a^2+b^2
hence;
c^2=5^2+12^2
c^2=144+25
c^2=169
thus;
c=sqrt169
c=13 units;
thus the perimeter of the rhombus will be:
P=L+L+L+L
P=13+13+13+13
P=52 units
33 Million dollars have been made.
Answer:
There is exactly one more real solution or there is exactly one more complex solution
Step-by-step explanation:
A quadratic equation is a polynomial of degree two
What this means is that a polynomial has two answers.
Now, from the question, we have an answer already which is a real root
Then the other answer which we do not have can take the form of two answers
It can either be a complex root or other wise be a real root
So the answer to this question is that ;
There is exactly one more real solution or there is exactly one more complex solution
As written, the denominator in both fractions is x, so the only restriction on the domain is ... x ≠ 0.
_____
We suspect you intend ...
... f(x) = 2/(x-4) +1/(x+2)
which is undefined when x = 4 or x = -2.
The domain is all real numbers except -2 and 4.
