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4 years ago

5

Frankie is practicing for a 5−kilometer race. His normal time is 31 minutes 25 seconds. Yesterday it took him only 29 minutes 38

seconds. How much faster was Frankie yesterday than his normal time?

1 answer:

gayaneshka [121]4 years ago

3 0

Frankie was 107 seconds faster or 1 minute and 47 seconds faster

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Solve 5h + 2(11 - h) = -5 (Find h. Explain each step please.)

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A store finds that its sales revenue changes at a rate given by S'(t) = −30t2 + 420t dollars per day where t is the number of da

allochka39001 [22]

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Step-by-step explanation:

Give the rate of change of sales revenue of a store modeled by the equation $S'(t)= -30t^{2} + 420t$ . The Total sales revenue function S(t) can be gotten by integrating the function given as shown;

$\int\limits {S'(t)} \, dt = \int\limits ({-30t^{2}+420t }) \, dt \\S(t) = \frac{-30t^{3} }{3}+\frac{420t^{2} }{2}\\ S(t)= -10t^{3} +210t^{2} \\$

a) The total sales for the first week after the campaign ends (t = 0 to t = 7) is expressed as shown;

$Given\ S(t) = -10t^{3} + 210t^{2}$

$S(0) = -10(0)^{3} + 210(0)^{2}\\S(0) = 0\\S(7) = -10(7)^{3} + 210(7)^{2}\\S(7) = -3430+10,290\\S(7) = 6,860$

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b) The total sales for the secondweek after the campaign ends (t = 7 to t = 14) is expressed as shown;

Total sales for the second week = S(14)-S(7)

Given S(7) = 6,860

To get S(14);

$S(14) = -10(14)^{3} + 210(14)^{2}\\S(14) = -27,440+41,160\\S(14) = 13,720$

The total sales for the second week after campaign ends = 13,720 - 6,860

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