At the time the cannonball reaches the ground, the vertical height of the cannonball from the ground is 0, i.e. h = 0.
Thus, the time it will take the cannonball to reach the ground is given by
Therefore, it will take the cannonball 8 seconds to reach the ground.
Lets approach this problem differently
15 gallons ....you drive ......330 miles lets divide both numbers by 15 to find out how much you can go with 1 gallon
1 gallon..... you drive.....330/15=22
you car goes 22 miles per 1 gallon
if you put 20 gallons
20*22=440 miles
Answer:
ratio of GEOM to TEYR is 
Therefore OE is equal to:
12 units is the length of OE, so the final answer is:
12.0 Units.
Answer:
2x+y=15x+y=10
Consider the first equation. Subtract 15x from both sides.
2x+y−15x=y
Combine 2x and −15x to get −13x.
−13x+y=y
Subtract y from both sides.
−13x+y−y=0
Combine y and −y to get 0.
−13x=0
Divide both sides by −13. Zero divided by any non-zero number gives zero.
x=0
Consider the second equation. Insert the known values of variables into the equation.
15×0+y=10
Multiply 15 and 0 to get 0.
0+y=10
Anything plus zero gives itself.
Assume that the amount needed from the 5% solution is x and that the amount needed from the 65% solution is y.
We are given that, the final solution should be 42 ml, this means that:
x + y = 42 ...........> equation I
This can also be written as:
x = 42-y .......> equation II
We are also given that the final concentration should be 45%, this means that:
5% x + 65% y = 45% (x+y)
0.05x + 0.65y = 0.45(x+y)
We have x+y = 42 from equation I, therefore:
0.05x + 0.65y = 0.45(42)
0.05x + 0.65y = 18.9 .........> equation III
Substitute with equation II in equation III as follows:
0.05x + 0.65y = 18.9
0.05(42-y) + 0.65y = 18.9
2.1 - 0.05y + 0.65y = 18.9
0.6y = 18.9 - 2.1
0.6y = 16.8
y = 28 ml
Substitute with y in equation II to get x as follows:
x = 42-y
x = 42 - 28
x = 14 ml
Based on the above calculations:
amount of 5% solution = x = 14 ml
amount of 65% solution = y = 28 ml
The correct choice is:
The teacher will need 14 mL of the 5% solution and 28 mL of the 65% solution.
