Answer:
Condition A.
A rectangle with four right angles
There can be many quadrilaterals satisfying this condition.
Condition B.
A square with one side measuring 5 inches
There can be only one quadrilateral satisfying this condition.
Condition C.
A rhombus with one angle measuring 43°
There can be many quadrilaterals satisfying this condition.
Condition D.
A parallelogram with one angle measuring 32°
There can be many quadrilaterals satisfying this condition.
Condition E.
A parallelogram with one angle measuring 48° and adjacent sides measuring 6 inches and 8 inches.
There can be only one quadrilateral satisfying this condition.
Condition F.
A rectangle with adjacent sides measuring 4 inches and 3 inches.
There can be only one quadrilateral satisfying this condition
Step-by-step explanation:
4/5y + 3(2/5x + 7/5y)...distribute thru the parenthesis
4/5y + 6/5x + 21/5y ...combine like terms
25/5y + 6/5x.....reduce
5y + 6/5x
Since division is the opposite of multiplication, you can turn this division problem into a multiplication problem by multiplying the top fraction by the reciprocal of the bottom.
