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Nastasia [14]
3 years ago
13

Laura is baking a cake. The recipe says that she has to mix 64 grams of chocolate powder to the flour. Laura knows that 1 cup of

that particular chocolate powder has a mass of 128 grams. She added two over three of a cup of chocolate powder to the flour. Should Laura add more chocolate powder to make the exact recipe or did she go over and by what amount?
She went over by one over three of a cup

She needs to add one over three of a cup

She needs to add one over six of a cup

She went over by one over six of a cup
Mathematics
1 answer:
pickupchik [31]3 years ago
5 0
The answer is <span>She went over by one over six of a cup
</span>
<span>1 cup of that particular chocolate powder has a mass of 128 grams: 1c = 128g

</span><span>64 grams of chocolate powder is x cups.
1c = 128g
x = 64g

1c : 128 g = x : 64g
x = </span>1c : 128 g * 64 g
x = 0.5 c
64 grams of chocolate powder is 0.5 cups = 1/2 cups

<span>She added two over three of a cup of chocolate powder: 2/3 cups

She need to add 1/2 cups: 1/2 = 3/6 cups
She added 2/3 cups: 2/3 = 4/6 cups

So she added 4/6 - 3/6 = 1/6 cups more.
</span>
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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
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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}

A = \dfrac{15}{2}

A = 7.5° C

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

y = 28.36^0 \ C    ( to two decimal places)

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