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d1i1m1o1n [39]
2 years ago
10

Decide, without calculation, if each of the integrals below are positive, negative, or zero. Let D be the region inside the unit

circle centered at the origin. Let T, B, R, and L denote the regions enclosed by the top half, the bottom half, the right half, and the left half of unit circle, respectively.1. ∬B xe^xdA2. ∬R xe^xdA3. ∬T xe^xdA4. ∬D xe^xdA5. ∬L xe^xdA? Positive Negative Zero
Mathematics
1 answer:
Schach [20]2 years ago
4 0

The integrals over B and T will be positive. Keeping y fixed, xe^x is strictly increasing over D as x increases, so the integrals over x (i.e. the bottom/top left quadrants of D) is negative but the integrals over x>0 are *more* positive.

The integrals over R and L are zero. If we take f(x,y)=xe^x, then f(x,-y)=f(x,y), which is to say f is symmetric across the x-axis. For the same reason, the integral over all of D is also zero.

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A scuba diver starts at 79.5 meters below the surface and descends until he reaches 108.3 meters below sea level. How many meter
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Answer:

28.8

Step-by-step explanation:

You are looking for the number between 108.3 and 79.5, so you have to do this 108.3-79.5

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2 years ago
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Jessica is 5 years older than her sister Jenna. Jenna tells her sister that in 5 years, she will be as old as Jessica was 5 year
emmasim [6.3K]
                          Jenna's age                   Jessica's age

today                         x                               x + 5


in 5 years                 x+5 

5 years ago                                                   x + 5 - 5


Equality                  x + 5 = x + 5 - 5

=> x + 5 = x


The equation is x + 5 = x


That equation has not any solution, because it reduces to 5 = 0, which is an absurd.


So, the conclusion is that information given by Jenna is wrong.


It is logical: if the difference of the ages is 5, in five years Jenna will have 10 years more than what her sister had 5 years ago, no 5 as she told Jessica.
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2 years ago
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A weather balloon has a volume of 90.0 L when it is released from sea level 101 kPa. What is the atmospheric pressure on the bal
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Answer:

<em><u>The final atmospheric pressure is 5.19 · 10⁴ Pa</u></em>

Step-by-step explanation:

Assuming that the temperature of the air does not change, we can use Boyle's law, which states that for a gas kept at constant temperature, the pressure of the gas is inversely proportional to its volume. In formula,

pV = const.

where p is the gas pressure and V is the volume

The equation can also be rewritten as

p₁ V₁ = p₂ V₂

where in our problem we have:

p₁ = 1.03 · 10₅ Pa is the initial pressure (the atmospheric pressure at sea level)

V₁ = 90.0L is the initial volume

p₂ is the final pressure

V₂ = 175.0L is the final volume

Solving the equation for p2, we find the final pressure:

p₂ = p₁ v₁ divided by V₂ = (1.01 · 10⁵)(90.0) divided by 175.0 = 5.19 · 10⁴ Pa

8 0
2 years ago
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Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist
GaryK [48]

Answer:

Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue; (4, 2)

For Broadway; (7.97, 2.49)

Step-by-step explanation:

Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit

The scale of the drawing is 1 unit = 0.25 miles

Therefore;

The actual length of the initial parade =0.25×12.265 unit = 3.09 miles

The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles

Part B:

For an actual length of 3 miles, the length on the scale drawing should be given as follows;

1 unit = 0.25 miles

0.25 miles = 1 unit

1 mile =  1 unit/(0.25) = 4 units

3 miles = 3 × 4 units = 12 units

With the same end point and route, we have;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

y² + (6 - x)² = 176 - 96·√2

y² = 176 - 96·√2 - (6 - x)²............(1)

Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2

Which gives;

y/x = 1/2

y = x/2 ..............................(2)

Equating equation (1) to (2) gives;

176 - 96·√2 - (6 - x)² = (x/2)²

176 - 96·√2 - (6 - x)² - (x/2)²= 0

176 - 96·√2 - (1.25·x²- 12·x+36) = 0

Solving using a graphing calculator, gives;

(x - 9.941)(x + 0.341) = 0

Therefore;

x ≈ 9.941 or x = -0.341

Since l₁ is required to be 12 - 4·√2, we have and positive, we have;

x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97

Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible

Part C: The coordinates of the cameras stationed half way down each road are;

For central avenue;

Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)

For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

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2 years ago
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