Help i dont understand
1 answer:
You might be interested in
The heights the balls hit a fence at 350 ft distance are 65 feet, 38 feet and 30 feet, respectively
<h3>Represent the distance-height relationship for each player’s ball as an equation, in a table and on a graph. </h3><u>Juan</u>
Juan's equation is given as:
h = -0.001d^2 + 0.5d + 2.5
h =
Set d to multiples of 50 from 0 to 400.
So, the table of values of Juan's function is:
d (ft) h(ft)
0 2.5
50 25
100 42.5
150 55
200 62.5
250 65
300 62.5
350 65
400 42.5
See attachment for the graph of Juan's function
<u>Mark</u>
A quadratic function is represented as:
h = ad^2 + bd + c
Using the values on the table of values, we have:
c = 3 -- the constant value
So, the equation becomes
h = ad^2 + bd + 3
Using the two other values on the table of values, we have:
23 = a(50)^2 + b(50) + 3
38 = a(100)^2 + b(100) + 3
Using a graphing tool, we have:
a = -0.001
b = 0.45
So, Mark's equation is h(d) = -0.001d^2 + 0.45d + 3
See attachment for Mark's graph.
<u>Barry</u>
From the graph, we have the table of values of Barry's function to be:
d (ft) h(ft)
0 2.5
50 21
100 35
150 44
200 48
250 46
300 41
350 30
400 14
450 0
Using a graphing tool, we have the quadratic function to be:
y = -0.001x^2 +0.4x +2.5
<h3><u>The shortest and the greatest distance before hitting the ground</u></h3>From the graphs, equations and tables, the distance travelled by the balls are:
Juan = 505 feet
Mark = 457 feet
Barry = 450 feet
This means that Juan's ball would travel the greatest distance while Barry's ball would travel the shortest.
<h3>The height the balls hit a fence at 350 ft distance</h3>To do this, we set d = 350
From the graphs, equations and tables, the height at 350 ft by the balls are:
Juan = 65 feet
Mark = 38 feet
Barry = 30 feet
The above represents the height the balls hit the fence
Read more about quadratic functions at:
#SPJ1
Answer:
Hence the maximum possible volume will be the 778.53 c.c
Step-by-step explanation:
Given:
A rectangle with 13 x 26 dimensions
And corners are cut to form side squares.
To Find:
Maximum possible volume for box
Solution :
Consider a rectangle of 13 x 26 dimension with and side of square at corner be x.
(Refer the attachment)
Now,
Formulating the volume equation for the box
So corner square sides we are going to fold up which makes height of the box
and remaining part will be length and breadth
As shown in fig,
Length=26-x
breadth=13-x
And height will be x
To get maximum volume differentiate the above equation,
Now ,Solve the Quadratic Equation to get x values,
=0
x=[-b±(b^2-4ac)^1/2]/2a
x=[78±Sqrt[(78)^2-4*338*3)]/2*3
x=[78±Sqrt(3028)]/6
x=[78±55.027]/6
x=78+55.027/6 or x=78-55.027/6
x=22.17 or x=3.8288
Use these values in 6x-78 to know which value posses the max and min value for the function.
So when x=22.17
6x-78=6*22.17-78
=55.02>0 i.e function will have minimum value .
When x=3.8288
6*3.8288-78
=-55.0272<0 i.e. Function will have maximum value
Now, the function will defines the maximum volume
