Answer:
See the explanation.
Step-by-step explanation:
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.
(Answer)
Answer:
Step-by-step explanation:
Given tq equation of the height reached by the rocket as y=-16x^2+235x+133
The velocity of the object at its maximum height is zero
v = dy/dx = 0(at max height)
v =-32x +235
0 =-32x+235
x = 235/32
x = 7.34
Hence the rocket will reach its maximum height after 7.34seconds
Answer:
24
Step-by-step explanation:
30 secs = 0.5 min
thus 84 / 3.5 min = 168 / 7 = 24
x and x+4 are the numbers
3x=2(x+4)
3x=2x+8
3x-2x=8
x=8
x+4=12
Answer:
y=8x-69
Step-by-step explanation:
you would use the equation y-y1 = m (x - x1)
plug in m = 8 , y1 = 3, x1 = 9
now you have y - 3 = 8 (x-9)
distribute 8 --> y-3 = 8x - 72
add three to both sides to get y = 8x - 69
