The fraction is 2/3
So, there should be 100 squares.
Next, the 100 squares is divided into 3 by vertical lines. The lines will not coincide with the sides of the squares. This is okay. Next, the 2 out of 3 will be shaded, this will be equivalent to 66 squares and one more square but it will not be filled up completely. In percent form, the fraction is 66.67%.
Area=height times base (for some prisms including cylinders)
vcylinder=hpir²
h=height
therefor pir²=base area
vcylinder=height times aeraofbase
given
h=9
v=324pi
324pi=9(basearea)
divide both sides by 9
36pi=areabase
the area of te base is 36pi square cm (put 36 in the blank since th pi is alredy there)
Answer:
kick 1 has travelled 15 + 15 = 30 yards before hitting the ground
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Step-by-step explanation:
1st kick travelled 15 yards to reach maximum height of 8 yards
so,
it has travelled 15 + 15 = 30 yards before hitting the ground
2nd kick is given by the equation
y (x) = -0.032x(x - 50)

we know that maximum height occurs is given as
and maximum height is

y = 20
so kick 2 travels 25 + 25 = 50 yards before hitting the ground
first kick reached 8 yards and 2nd kick reached 20 yards
Cows : sheep = 5 : 4
5x + 4x = 648
9x = 648
x = 72
5x = 5 * 72 = 360
4x = 4 * 72 = 288
There are originally 360 cows and 288 sheep.
2/3 of the sheep are sold. That means that 1/3 of the sheep are left.
1/3 * 288 = 96
96 sheep are left.
x cows were sold.
(360 - x)/96 = 3/1
360 - x = 288
-x = -72
x = 72
72 cows were sold
Answer:
Two distinct concentric circles: 0 max solutions
Two distinct parabolas: 4 max solutions
A line and a circle: 2 max solution
A parabola and a circle: 4 max solutions
Step-by-step explanation:
<u>Two distinct concentric circles:</u>
The maximum number of solutions (intersections points) 2 distinct circles can have is 0. You can see an example of it in the first picture attached. The two circles are shown side to side for clarity, but when they will be concentric, they will have same center and they will be superimposed. So there can be ZERO max solutions for that.
<u>Two distinct parabolas:</u>
The maximum solutions (intersection points) 2 distinct parabolas can have is 4. This is shown in the second picture attached. <em>This occurs when two parabolas and in perpendicular orientation to each other. </em>
<u>A line and a circle:</u>
The maximum solutions (intersection points) a line and a circle can have is 2. See an example in the third picture attached.
<u>A parabola and a circle:</u>
The maximum solutions (intersection points) a parabola and a circle can have is 4. If the parabola is <em>compressed enough than the diameter of the circle</em>, there can be max 4 intersection points. See the fourth picture attached as an example.