9514 1404 393
Answer:
671 feet
Step-by-step explanation:
There are a couple of ways to figure this. One is to use a sort of shortcut equation to find the distance traveled (d) by an object when subject to some initial velocity (v) and acceleration (a). Here the acceleration due to gravity is -32 ft/s².
v² = 2ad
d = v²/(2a) = (192 ft/s)^2/(2·32 ft/s²) = 576 ft
This height is in addition to the starting height of 95 ft, so the arrow's maximum height is ...
max height = 95 ft + 576 ft = 671 ft
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Another way to work this problem is to start with the equation for ballistic motion. Filling in the given initial velocity and height, we have ...
h(t) = -16t^2 +192t +95
The time the arrow reaches the maximum height is the time representing the axis of symmetry of the parabola:
t = -(192)/(2(-16)) = 6
Then the maximum height is ...
h(6) = -16·6^2 +192·6 +95 = 671
The maximum height is 671 feet.
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<em>Additional comment</em>
For the standard-form quadratic ...
y = ax^2 +bx +c
The axis of symmetry is ...
x = -b/(2a)
Area for circle is πr² so πr² =28.26 and we can sub 3.14 as pi
3.14*r²=28.26 we can divide by 3.14 to get r² on it's own
r²=28.26/3.14
then we root both sides to get r on it's own
28.26/3.14=9 √9=3
and the diameter is double the radius 3*2=6 so the diameter is 6
To get your answer:
9 times 5 times two= 90
9 times 7.5 times two= 135
5 times 7.5 times two= 75
then add all of them together and you will get the surface area
90+135+75=300
you should have an answer of 300 feet²
Answer:
The value of the account in the year 2009 will be $682.
Step-by-step explanation:
The acount's balance, in t years after 1999, can be modeled by the following equation.
In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.
$330 in an account in the year 1999
This means that 
$590 in the year 2007
2007 is 8 years after 1999, so P(8) = 590.
We use this to find r.
Applying ln to both sides:
Determine the value of the account, to the nearest dollar, in the year 2009.
2009 is 10 years after 1999, so this is A(10).
The value of the account in the year 2009 will be $682.
400 children and 600 adults bought tickets.
Step-by-step explanation:
Given,
Cost of one child ticket = $14
Cost of one adult ticket = $32
Total attendance = 1000
Revenue generated = $24800
Let,
x be the number of children.
y be the number of adults.
According to given statement;
x+y=1000 Eqn 1
14x+32y=24800 Eqn 2
Multiplying Eqn 1 by 14
Subtracting Eqn 3 from Eqn 2
Dividing both sides by 18
Putting y=800 in Eqn 1
400 children and 600 adults bought tickets.
Keywords: linear equations, subtraction
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