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
Answer:
x=9
Reverse. 14-5=9, so x must be 9.
These are two separate problems: in the first we will have to substitute in a new value for x into the original equation and in the second we will manipulate the preexisting equation for f(x).
To begin, we will sub in f(x/3). To do this, we will substitute each variable x in the equation (in this case there is only one) with x/3, and then simplify the resulting equation.
f(x) = 6x - 18
f(x/3) = 6(x/3) - 18
To simplify, we should distribute the 6 on the right side of the equation.
f(x/3) = 6x/3 - 18
Now, we can divide the first term on the right side to finalize our simplification.
f(x/3) = 2x -18
Secondly, we are asked to find f(x)/3. To do this, we will take our original value for f(x), and then simplify divide that entire function by 3.
f(x) = 6x - 18
f(x)/3 = (6x-18)/3
This means that we must divide each term of the binomial by 3, so we are really computing
f(x)/3 = 6x/3 - 18/3
We can simplify by dividing both of the terms.
f(x)/3 = 2x - 6
Therefore, your answer is that f(x/3) = 2x - 18, but f(x)/3 = 2x - 6. It is important to recognize that these are two similar, yet different, answers.
Hope this helps!
