F(3)-f(2)=-2(x-3)2+4
2 answers:
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Answer:
Part a) The number of boxes needed will be 4
part b) The fraction of the box that is empty is
Part c) Are needed 2 cupcakes to fill the last box
Step-by-step explanation:
Part a) How many boxes will be needed for Polly to bring all the cupcakes to the party?
To find out the number of boxes needed, divide the total number of cupcakes by 4 (maximum number of cupcakes that fit in each box).
so
Round up
The number of boxes needed will be 4
Part b) If the bakery completely fills as many boxes as possible, what fraction of the last box is empty?
we know that
The maximum number of cupcakes per box is 4
The first three boxes have 4 cupcakes
The last box has
The fraction of the box that is empty is equal to the number of cupcakes missing to fill the box divided by the cupcake capacity of the box
Part c) How many more cupcakes are needed to fill this box
Subtract the number of cupcakes in the box from the cupcake capacity of the box
The last box has 2 cupcakes
The cupcake capacity of the box is 4
so
therefore
Are needed 2 cupcakes to fill the last box
Define
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s