Consider the vertices of parallelogram JKLM with vertices J(2,2) , K(5,3) , L(5,-3) and M(2,-4).
Perimeter JKLM = Length JK + Length KL + Length LM + Length JM
Length JK = (2,2) (5,3)
The length(or distance) between two points say 
and 
is given by the distance formula:
Now, length JK = 
= 
units
Since, JKLM is a parallelogram. In parallelogram opposite sides are equal in length.
Therefore, LM = units
Now, length KL = 
= 6 units
Since, JKLM is a parallelogram. In parallelogram opposite sides are equal in length.
Therefore, JM = 6 units
Perimeter of JKLM = + + 6 + 6
= 2 + 12
= 18.324
Rounding to the nearest tenth, we get
= 18.3 units.
Therefore, the perimeter of JKLM is 18.3 units.
Hello,
the transformation that applies (x,y) to (-x,-y) is a symmetry by the origin.
Every symmetry transforms a line into a line.
So the new points are also collinear.
THIS ANSWER IS FOR f(3):
f(x)=x+3
f(3)=3+3
f(3)=6
THIS ANSWER IS FOR f(-2):
f(x)=x+3
f(-2)=-2+3
f(-2)=1
Angle 3 is 99 degrees.
Angle 2 is 61 degrees.
angle 1 is also 61 degrees.
2^2x=5^x−1
Take the log pf both sides:
ln(2^2x) = ln(5^x-1)
Expand the logs by pulling the exponents out:
2xln(2) = (x-1)ln(5)
Simpligy the right side:
2xln(2) = ln(5)x - ln(5)
Now solve for x:
Subtract ln(5)x from both sides:
2xln(2) - ln(5)x = -ln(5)
Factor x out of 2xln(2)-ln(5)x
x(2ln(2) - ln(5)) = -ln(5)
Divide both sides by (2ln(2) - ln(5))
X = - ln(5) / (2ln(2) - ln(5))
