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Lapatulllka [165]
4 years ago
12

Alejandra can husk 8 ears of corn in 24 minutes. At this rate, how many ears of corn can she husk in 33 minutes?

Mathematics
2 answers:
Lesechka [4]4 years ago
7 0

Answer:

11

Step-by-step explanation:

8_murik_8 [283]4 years ago
5 0
I'm afraid you're wrong. We may divide 24 by 8 to find out how long does it take for Alejandra to husk 1 ear of corn and then just divide 33 by the result:

24 ÷ 8 = 3

So Alejandra husks 1 ear of corn in 3 minutes.

33 ÷ 3 = 11

Answer: In 33 minutes Alejandra can husk <u>11 ears of corn</u>.
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The daily average temperature in Santiago, Chile, varies over time in a periodic way that can be modeled approximately by a trig
natali 33 [55]

Answer:

a) the trigonometric function is;

y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

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Step-by-step explanation:

This data can be represented by the sinusoidal function of the form :

\mathbf{y = A sin (Bt -C)+D}

where A = amplitude and which can be determined via the formula:

A = \dfrac{largest \ temperature -  lowest \ temperature}{2}

A = \dfrac{29-14}{2}

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where B = the frequency;

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B = \dfrac{2 \pi}{365}   ( where 365 is the time period )

The vertical shift is found by the equation D;

D =  \frac{largest \ temperature + lowest \ temperature}{2}

D = \frac{29+14}{2}

D = 21.5

Replacing the values of A ; B and D into the above sinusoidal function; we have :

y = 7.5 sin (\frac{2 \pi}{365}t -C) + 21.5

From the question; when it is 7th of the year ( i.e January 7);

t =  7 and the temperature (y) = 29° C

replacing that too into the above equation; we have:

29= 7.5 sin (\frac{2 \pi}{365}*7 -C) + 21.5

29= 7.5 sin (\frac{14 \pi}{365} -C) + 21.5

\frac{29-21.5}{7.5}=  sin (\frac{14 \pi}{365} -C)

1=  sin (\frac{14 \pi}{365} -C)

sin^{-1}(1)=   (\frac{14 \pi}{365} -C)

\frac{\pi}{2}=   (\frac{14 \pi}{365} -C)

C=   (\frac{14 \pi}{365} -\frac{\pi}{2})

C=   (\frac{28 \pi- 365 \pi}{730} )

C=  \frac{-337 \pi}{730}

Thus; the trigonometric function is;

y = 7.5 sin ( \frac{2 \pi}{365}t + \frac{337 \pi}{730})+ 21.5

Similarly; to determine the temperature o Jan 31; i.e when t= 31 ; we have :

y = 7.5 sin ( \frac{2 \pi}{365}*31+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{62 \pi}{365}+ \frac{337 \pi}{730})+ 21.5

y = 7.5 sin ( \frac{124 \pi+ 337 \pi }{730})+ 21.5

y = 7.5 sin ( \frac{461 \pi }{730})+ 21.5

y = 7.5 *( 0.915)+ 21.5

y = 6.8689+ 21.5

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