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dolphi86 [110]
3 years ago
13

Given the function F (x) = x2 -2x -5 determine the average rate of change of the function over the interval -5 < x <6.

Mathematics
1 answer:
KengaRu [80]3 years ago
4 0

Answer:

1

Step-by-step explanation:

Here f(x) = x^2 - 2x - 5, which at x - -5 is 25 +10 - 5 = 30 and at x = 6 is 19.

The average value of a function f(x) over an interval [a, b] is

                  f(b) - f(a)

ave. val. = ---------------

                     b - a

which in this particular case is

                     19 - 30

ave. val. = ----------------- = -11/11 = 1

                     6 - (-5)

The average value of this function f(x) = x^2 - 2x - 5 on [-5, 6] is 1.

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What is the equation of the line connecting (1,5) and (4,14)?
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To find the slope of the line:
(y2-y1)/(x2-x1)
(14-5)/(4-1)=(9)/(3)=3

Only one of your 4 possible answers has 3 as a slope. However, plugging in each point into the y=mx+b equation, the y-intercept consistently comes out as 2..
y=mx+b
14=3(4)+b  b=2
5=3(1)+b    b=2
y=3x+2
If there is a consistent, positive slope (from your question, this does not seem to have a quadratic as an option), 3x+5 is not even a viable solution because x=1 when the y-value is 5 (and thus no other x value {0} could have a y-value of 5). It seems as though you have a typo on your hands. Hopefully this helps?
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4 years ago
Nine times a number, diminished by 4, is 95
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Answer:

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Luiza is jumping on a trampoline.
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Answer:

The second time when Luiza reaches a height of 1.2 m = 2 08 s

Step-by-step explanation:

Complete Question

Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.

H(t) = -0.6 cos (2pi/2.5)t + 1.5.

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Solution

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H(t) = -0.6cos⁡(2π/2.5)t + 1.5

What is t when H = 1.2 m

1.2 = -0.6cos⁡(2π/2.5)t + 1.5

0.6cos⁡(2π/2.5)t = 1.2 - 1.5 = -0.3

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Note that in radians,

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Cos (5π/3) = 0.5

So,

Cos (2π/2.5)t = Cos (5π/3)

(2π/2.5)t = (5π/3)

(2/2.5) × t = (5/3)

t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.

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