H= (2pier^2)/ 2pier all subtracted by s so the last equation from the picture.
Hope it helps!
I=k/d^2
4=k/d^2 and 1=k/64 so if we divide the first by the second we get:
4/1=(k/d^2)/(k/64)
4=(k/d^2)*(64/k)
4=64/d^2
d^2=64/4
d^2=16
d=4 meters
A quadrilateral is any figure with 4 sides, no matter what the lengths of
the sides or the sizes of the angles are ... just as long as it has four straight
sides that meet and close it up.
Once you start imposing some special requirements on the lengths of
the sides, or their relationship to each other, or the size of the angles,
you start making special kinds of quadrilaterals, that have special names.
The simplest requirement of all is that there must be one pair of sides that
are parallel to each other. That makes a quadrilateral called a 'trapezoid'.
That's why a quadrilateral is not always a trapezoid.
Here are some other, more strict requirements, that make other special
quadrilaterals:
-- Two pairs of parallel sides . . . . 'parallelogram'
-- Two pairs of parallel sides
AND all angles the same size . . . . 'rectangle'
(also a special kind of parallelogram)
-- Two pairs of parallel sides
AND all sides the same length . . . 'rhombus'
(also a special kind of parallelogram)
-- Two pairs of parallel sides
AND all sides the same length
AND all angles the same size . . . . 'square'.
(also a special kind of parallelogram, rectangle, and rhombus)
Answer:
32y^2-32
Step-by-step explanation:
4y^2+4(7y^2-8)
4y^2+28y^2-32
32y^2-32
if you need extra steeps explaining then plz let me know and I will help :)
Easy the answer is B: 17 and one-third
