Given:
2 parallelograms with an area of 9 1/3 yd²
height of each parallelogram is 1 1/3 yd
Area of parallelogram = base * height
We need to divide the combined area into two to get each parallelogram's base.
9 1/3 = ((9*3)+1)/3 = 28/3
28/3 ÷ 2 = 28/3 * 1/2 = 28/6 yd² or 4 4/6 yd² ⇒ 4 2/3 yd²
Area of each parallelogram is 4 2/3 yd²
4 2/3 yd² = base * 1 1/3 yd
14/3 yd² ÷ 4/3 yd = base
14/3 yd² x 3/4 yd = base
14*3 / 3*4 = base
42 / 12 = base
3 6/12 yd = base
or 3 1/2 yd = base
a) the base of each parallelogram is 3 1/2 yards
b) we can assume that the two parallelograms form a rectangle.
area of a rectangle is length times width.
length is 3 1/2 yds * 2 = 7 yds
width is 3 1/2 yds
Area of rectangle = 7 yds * 3 1/2 yds
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = 24 1/2 yd²
1st worker produced x pants,
2nd worker produced (x+8) pants.
x+x+8=86
2x=78
x= 39
x+8 = 39+8 = 47
1st worker produced 39 pants,
2nd worker produced 47 pants.
Answer:
L(w) = 8 mm + (2 mm/wk)(wk)
Step-by-step explanation:
L(w) = length of beard as a function of time in weeks
L(w) = 8 mm + (2 mm/wk)(wk)
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
