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kotegsom [21]
3 years ago
13

Mrs. kirby, 58, is a moderately active woman. she works in the garden almost daily, occasionally walks with her friends, and bab

ysits her three very active young grandchildren every weekend. she eats a fairly standard diet every day: she has cereal with milk and coffee for breakfast, some kind of salad for lunch, and a piece of meat or fish with a healthy of veggies for dinner. she usually skips desserts or has just a couple of strawberries. when asked to look at the food labels and calculate her intake of carbohydrates, fats, and proteins in grams over the entire day, on an average day, she came up with the following numbers: carbohydrates: 243 g fats: 41 g proteins: 63 g calculate what percentage of mrs. kirby's total energy intake came from carbohydrates.
Mathematics
1 answer:
uranmaximum [27]3 years ago
5 0

Answer:

61,01%

Step-by-step explanation:

We must consider that her total energy intake is the addition of the energy given by the carbohydrates, fats and proteins she takes.

1g of carbohydrates contains 4kcal

1g of proteins contains 4kcal

1g of fat contains 9 kcal.

That is: 243 g of carbohidrates is equal to 972 kcal; 41g of fat is equal to 369 kcal and 63 g of proteins is equal to 252 kcal.

If we sum up all the calories she intakes in one day, the result is= 972kcal+369kcal+252kcal=1593kcal.

Now, we make a rule of 3: If 1593kcal is the 100% of the intake, then 972kcal = 972kcal*100%/1593kcal = 61,01%

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Find the area of the region that lies inside the first curve and outside the second curve.
marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the  the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = \dfrac{1}{2}

\theta = -\dfrac{\pi}{3}, \ \  \dfrac{\pi}{3}

Now, the area of the  region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \  \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \  5^2 d \theta

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A = 25 \begin{bmatrix}   \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3}   \end {bmatrix}- \dfrac{ 25 \pi}{3}

A =    \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx
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3 years ago
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