We want to put b alone on one side of the equation.
If we multiply by 1/2, we get 1/4bh = A/2. Not only did we not remove a variable from the left side, we made it even more complicated.
So the answer is A.
Multiples of 30 : 30,60,90,120,150....
and the first one that 4 goes into evenly is 60
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:
First option
Step-by-step explanation:
Have a nice day
hi,
you must replace x by the number between parenthese.
I show you with the first one and let you do the second one
p(x) = 11x^5 -11x^4 - 5x^2 +15x-8
p(-4) = 11 (-4)^5 - 11 (-4)^4 -5(-4)² +15(-4) -8
p(-4) = 11 ( -1024- 256) - 5*16 -60-8
p(-4) = 11 ( -1280) -80-60-8
p(-4) = - 14080 - 148
p(-4) = - 14 228