What is 1/3 of 12=2/3 of?
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<span>Given that the railroad rail is 6 kilometers before expansion.
When the railroad expands, it makes an isosceles triangle at the center of the rail.
The base of the triangle is 6 kilometers while each leg of the triangle is given by (6 + 10/100,000) / 2 = 6.0001 / 2 = 3.00005 kilometers.
[recall that 100,000 centimeters gives 1 kilometers, thus 10 centimetres gives 10/100,000 kilometers]
Taking one half of the isoceles triangle, we have a right triangle with one of the leg as 6 / 2 = 3 kilometers and the hypothenus = 3.00005 kilometers.
Thus, using the Pythagoras theorem, we find the other leg (height) of the right triangle as follows.
Let the other leg of the right triangle be h, then
Thus, the center of the rail will rise 0.01732058 kilometers above the ground.
But, there are 1,000 meters in 1 kilometers, therefore, </span><span><span>the center of the rail will rise 0.01732058 x 1,000 = 17.32 meters above the ground.</span></span>
When the railroad expands, it makes an isosceles triangle at the center of the rail.
The base of the triangle is 6 kilometers while each leg of the triangle is given by (6 + 10/100,000) / 2 = 6.0001 / 2 = 3.00005 kilometers.
[recall that 100,000 centimeters gives 1 kilometers, thus 10 centimetres gives 10/100,000 kilometers]
Taking one half of the isoceles triangle, we have a right triangle with one of the leg as 6 / 2 = 3 kilometers and the hypothenus = 3.00005 kilometers.
Thus, using the Pythagoras theorem, we find the other leg (height) of the right triangle as follows.
Let the other leg of the right triangle be h, then
Thus, the center of the rail will rise 0.01732058 kilometers above the ground.
But, there are 1,000 meters in 1 kilometers, therefore, </span><span><span>the center of the rail will rise 0.01732058 x 1,000 = 17.32 meters above the ground.</span></span>
25a. You are given a formula for height and told that the value of "r" in the formula is 19.6. You are asked to find the value of t such that h = 14.7. Substitute the given values and solve for t the way you solve any quadratic equation.
14.7 = 19.6t -4.9t²
4.9t² -19.6t +14.7 = 0 . . . . . put the equation in standard form
4.9(t² -4t +3) = 0 . . . . . . . . . factor out 4.9 to simplify the numbers
(t -1)(t -3) = 0 . . . . . . . . . . .. divide by 4.9 and factor
The solutions to this are ...
t = 1, t = 3
After 1 second, the ball will reach the height of 14.7 meters.
25b. This question asks you to find the value of t that make h = 0.
0 = 19.6t -4.9t² . . . . . . substitute the given numbers
0 = 4.9t(4 -t) . . . . . . . . factor
The solutions to this are ...
t = 0, t = 4
The ball will hit the ground after 4 seconds.
_____
As you can see from the graph below, a graphing calculator can be helpful for solving problems of this sort.
14.7 = 19.6t -4.9t²
4.9t² -19.6t +14.7 = 0 . . . . . put the equation in standard form
4.9(t² -4t +3) = 0 . . . . . . . . . factor out 4.9 to simplify the numbers
(t -1)(t -3) = 0 . . . . . . . . . . .. divide by 4.9 and factor
The solutions to this are ...
t = 1, t = 3
After 1 second, the ball will reach the height of 14.7 meters.
25b. This question asks you to find the value of t that make h = 0.
0 = 19.6t -4.9t² . . . . . . substitute the given numbers
0 = 4.9t(4 -t) . . . . . . . . factor
The solutions to this are ...
t = 0, t = 4
The ball will hit the ground after 4 seconds.
_____
As you can see from the graph below, a graphing calculator can be helpful for solving problems of this sort.