Simplify. Rewrite the expression in the form 4^n (4^9 ) / ( 4^5)
1 answer:
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9514 1404 393
Answer:
a) 2.038 seconds
b) 5.918 meters
c) 1.076 seconds
Step-by-step explanation:
For the purpose of answering these questions, it is convenient to put the given equation into vertex form.
h = -4.9t² +9.2t +1.6
= -4.9(t² -(9.2/4.9)t) +1.6
= -4.9(t² -(9.2/4.9)t +(4.6/4.9)²) +1.6 +4.9(4.6/4.9)²
= -4.9(t -46/49)² +290/49
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a) To find h = 0, we solve ...
0 = -4.9(t -46/49)² +290/49
290/240.1 = (t -46/49)² . . . . subtract 290/49 and divide by -4.9
√(2900/2401) +46/49 = t ≈ 2.0378 . . . . seconds
The ball takes about 2.038 seconds to fall to the ground.
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b) The maximum height is the h value at the vertex of the function. It is the value of h when the squared term is zero:
290/49 m ≈ 5.918 m
The maximum height of the ball is about 5.918 m.
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c) We want to find t for h ≥ 4.5.
h ≥ 4.5
-4.9(t -46/49)² +290/49 ≥ 4.5
Subtracting 290/49 and dividing by -4.9, we have ...
(t -46/49)² ≥ 695/2401
Taking the square root, and adding 46/49, we find the time interval to be ...
-√(695/2401) +46/49 ≤ t ≤ √(695/2401) +46/49
The difference between the interval end points is the time above 4.5 meters. That difference is ...
2√(695/2401) ≈ 1.076 . . . . seconds
The ball is at or above 4.5 meters for about 1.076 seconds.
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I like a graphing calculator for its ability to answer these questions quickly and easily. The essentials for answering this question involve typing a couple of equations and highlighting a few points on the graph.
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<em>Additional comment</em>
I have a preference for "exact" answers where possible, so have used fractions, rather than their rounded decimal equivalents. The calculator I use deals with these fairly nicely. Unfortunately, the mess of numbers can tend to obscure the working.
"Vertex form" for a quadratic is ...
y = a(x -h)² +k . . . . where the vertex is (h, k) and 'a' is a vertical scale factor.
In the above, we have 'a' = -4.9, and (h, k) = (46/49, 290/49) ≈ (0.939, 5.918)
As the mapping suggests, the first (x) coordinate is decreased by 3, and the second (y) coordinate is increased by 2. So, given the inputs, you have
And yes, this rule is a function: it has a starting set (the domain), which is all possible couples , and a landing set (the codomain), which is again composed by the couples
. Since for every point in the domain you have one and only one output, this rule is a function.