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statuscvo [17]
3 years ago
5

Sandi tracks her calories burned during water aerobics class. The number of calories she burns is expressed by the function c(t)

= 350t, where t is the number of hours spent doing water aerobics.
To burn more calories, Sandi wears ankle weights during her class. The number of calories she burns while wearing ankle weights is expressed by the function b(c) = 1.2c, where c is the number of calories burned doing water aerobics without weights.

Which of the following composite functions expresses the calories, as a function of time, Sandi burns while doing water aerobics with ankle weights?

A b[c(t)] = 351.2t A B C OR D?
B b[c(t)] = 420t
C c[b(t)] = 351.2t
D c[b(t)] = 420t
Mathematics
1 answer:
tigry1 [53]3 years ago
6 0
We have c(t) = 350t and b(c) = 12c

The composite function that expresses the calories as function of time is given by

b[c(t)] = 12[350t]
b[c(t)] = 420t

Correct answer: B
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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
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Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

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Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

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(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

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f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

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x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

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