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2 years ago

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An average person in the country needs 2,450 kilocalories(kcal) / day of food to meet their full nutritional needs. The average

number of kcal per hectare(ha) produced from the available food in the country is 15,737,000 kcal/ha
CALCULATE the amount of land that would be needed to produce enough kilocalories to feed a person for a year in this country. Show your work

1 answer:

OlgaM077 [116]2 years ago

7 0

Answer:

The answer to your question is 0.057 ha

Step-by-step explanation:

Data

Kilocalories needed for a person per day = 2450

Kilocalories produced per hectare = 15737000

Process

1.- Calculate the number of kcal needed per a person for a year

2450 kcal ----------------- 1 day

x ----------------- 365 days

x = (365 x 2450) / 1

x = 894250 kcal

2.- Calculate the amount of land needed

15737000 kcal ------------- 1 ha

894250 kcal ------------ x

x = (894250 x 1) / 15737000

x = 0.057 ha

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Let D be the smaller cap cut from a solid ball of radius 8 units by a plane 4 units from the center of the sphere. Express the v

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Answer:

Step-by-step explanation:

The equation of the sphere, centered a the origin is given by $x^2+y^2+z^2 = 64$ . Then, when z=4, we get

$x^2+y^2= 64-16 = 48$ .

This equation corresponds to a circle of radius $4\sqrt[]{3}$ in the x-y plane

c) We will use the previous analysis to define the limits in cartesian and polar coordinates. At first, we now that x varies from $-4\sqrt[]{3}$ up to $4\sqrt[]{3}$ . This is by taking y =0 and seeing the furthest points of x that lay on the circle. Then, we know that y varies from $-\sqrt[]{48-x^2}$ and $\sqrt[]{48-x^2}$ , this is again because y must lie in the interior of the circle we found. Finally, we know that z goes from 4 up to the sphere, that is , z goes from 4 up to $\sqrt[]{64-x^2-y^2}$

Then, the triple integral that gives us the volume of D in cartesian coordinates is

$\int_{-4\sqrt[]{3}}^{4\sqrt[]{3}}\int_{-\sqrt[]{48-x^2}}^{\sqrt[]{48-x^2}} \int_{4}^{\sqrt[]{64-x^2-y^2}} dz dy dx$ .

b) Recall that the cylindrical coordinates are given by $x=r\cos \theta, y = r\sin \theta,z = z$ , where r corresponds to the distance of the projection onto the x-y plane to the origin. REcall that $x^2+y^2 = r^2$ . WE will find the new limits for each of the new coordinates. NOte that, we got a previous restriction of a circle, so, since $\theta[\tex] is the angle between the projection to the x-y plane and the x axis, in order for us to cover the whole circle, we need that [tex]\theta$ goes from 0 to $2\pi$ . Also, note that r goes from the origin up to the border of the circle, where r has a value of 4\sqrt[]{3}. Finally, note that Z goes from the plane z=4 up to the sphere itself, where the restriction is \sqrt[]{64-r^2}. So, the following is the integral that gives the wanted volume

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta$ . Recall that the r factor appears because it is the jacobian associated to the change of variable from cartesian coordinates to polar coordinates. This guarantees us that the integral has the same value. (The explanation on how to compute the jacobian is beyond the scope of this answer).

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d ) Let us use the integral in cylindrical coordinates

$\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} \int_{4}^{\sqrt[]{64-r^2}} rdz dr d\theta=\int_{0}^{2\pi}\int_{0}^{4\sqrt[]{3}} r (\sqrt[]{64-r^2}-4) dr d\theta=\int_{0}^{2\pi} d \theta \cdot \int_{0}^{4\sqrt[]{3}}r (\sqrt[]{64-r^2}-4)dr= 2\pi \cdot (-2\left.r^{2}\right|_0^{4\sqrt[]{3}})\int_{0}^{4\sqrt[]{3}}r \sqrt[]{64-r^2} dr$

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Answer:

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Step-by-step explanation:

Combine like terms in 2x^2 +3 x-7 =x^2 +5x +39:

2x^2 - x^2 = x^2 (first term);

3x - 5x = -2x (second term);

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Then we have, all on the left side, x^2 - 2x - 46, which is a quadratic equation. Here the coefficients are a = 1, b = -2 and c = -46.

Then the discriminant, b^2 - 4ac, is:

(-2)^2 - 4(1)(-46) = 4 + 184.

The roots are:

-(-2) ±√188 2 ± √4√47

------------------- = -------------------

2 2

= 1 ± √47 (last of the answer choices)

=

x = ------------------

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