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)
Mike's withdrawals will be treated as negative numbers.
Since Mike withdrew $60 three times, use the following equation to find the change in his balance:
The change in Mike's account balance will be
-$180.
Answer:
The order of ordered pairs of a function and its inverse reverse. The graph of the inverse of this function passes through the points (2,-1), (6,0), (-4,3).
Explanation:
I just took the test.
change yrds to ft
5yds = 5*3 = 15 ft
5yds 2 ft = 15ft + 2ft = 17 ft
6yds = 6 * 3 = 18 ft
6 yds 1 ft = 18 + 1
19 ft
A = l * w
A = 17* 19
A = 323 ft^2
Choice A
Answer:
The values of x and y in the diagonals of the parallelogram are x=0 and y=5
Step-by-step explanation:
Given that ABCD is a parallelogram
And segment AC=4x+10
From the figure we have the diagonals AC=3x+y and BD=2x+y
By the property of parallelogram the diagonals are congruent
∴ we can equate the diagonals AC=BD
That is 3x+y=2x+y
3x+y-(2x+y)=2x+y-(2x+y)
3x+y-2x-y=2x+y-2x-y
x+0=0 ( by adding the like terms )
∴ x=0
Given that segment AC=4x+10
Substitute x=0 we have AC=4(0)+10
=0+10
=10
∴ AC=10
Now (3x+y)+(2x+y)=10
5x+2y=10
Substitute x=0, 5(0)+2y=10
2y=10
∴ y=5
∴ the values of x and y are x=0 and y=5
