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nataly862011 [7]
3 years ago
8

Will mark brainliest !! Easy math

Mathematics
1 answer:
marshall27 [118]3 years ago
6 0

Answer:

increase = 37.38 cm²

You'd like the first way because it took less effort and less chance of a miscalculation.

Step-by-step explanation:

first way:

(just work with the difference in height)

(0.7 x 8.7 x 2) + (0.7 x 18 x 2) = 37.38 cm²

second way:

old total lateral surface area = (31 x 8.7 x 2) + (31 x 18 x 2) = 1655.4

new total lateral surface area = (31.7 x 8.7 x 2) + (31.7 x 18 x 2) = 1692.78

increase = 1692.78 - 1655.4 = 37.38 cm²

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a. Note that f(x)=x^ne^{-2x} is continuous for all x. If f(x) attains a maximum at x=3, then f'(3) = 0. Compute the derivative of f.

f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}

Evaluate this at x=3 and solve for n.

n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0

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\dfrac n2 = \dfrac{3^n}{3^{n-1}}

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To ensure that a maximum is reached for this value of n, we need to check the sign of the second derivative at this critical point.

f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0

The second derivative at x=3 is negative, which indicate the function is concave downward, which in turn means that f(3) is indeed a (local) maximum.

b. When n=4, we have derivatives

f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}

Inflection points can occur where the second derivative vanishes.

12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0

12x^2 - 16x^3 + 4x^4 = 0

4x^2 (3 - 4x + x^2) = 0

4x^2 (x - 3) (x - 1) = 0

Then we have three possible inflection points when x=0, x=1, or x=3.

To decide which are actually inflection points, check the sign of f'' in each of the intervals (-\infty,0), (0, 1), (1, 3), and (3,\infty). It's enough to check the sign of any test value of x from each interval.

x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0

x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0

x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0

x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0

The sign of f'' changes to either side of x=1 and x=3, but not x=0. This means only \boxed{x=1} and \boxed{x=3} are inflection points.

4 0
1 year ago
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you can tell if fractions are proportional by using cross multiplication.
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