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jekas [21]
3 years ago
9

Heyyyyy wyddddddddddddddddddddddd

Advanced Placement (AP)
2 answers:
kvv77 [185]3 years ago
6 0
Failing my geo class
ANEK [815]3 years ago
3 0
nothinggggg hbuuuu bestieeee
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Explain the relationship of the fovea to cones in the retina?
Ira Lisetskai [31]

Rods and cones convert light to neural signals which the nervous system can use. ... The 100 million rods are locates in the retina away from the fovea, so they carry out peripheral vision ("side" vision towards the edge of the visual field). Figure 6 shows the distribution of rods and cones on the retina.



5 0
2 years ago
Match each type of tax to its description.
Furkat [3]
Sales tax is regressive, the second is progressive, and the third is proportional.
5 0
3 years ago
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The regions bounded by the graphs of y=x2 and y=sin2x are shaded in the figure above. What is the sum of the areas of the shaded
Alex Ar [27]

Answer:

The sum of the area of the shaded regions = 0.248685

Explanation:

The sum of the area of the shaded region is given as follows;

The point of intersection of the graphs are;

y = x/2

y = sin²x

∴ At the intersection, x/2 = sin²x

sinx = √(x/2)

Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

1/2 × (0.55)×0.55/2 ≈ 0.075625

The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

\int\limits {sin^n(x)} \, dx = -\dfrac{1}{n} sin^{n-1}(x) \cdot cos(x) + \dfrac{n-1}{n} \int\limits {sin^{n-2}(x)} \, dx

Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

∴ A₂ =1/2 × ((0.55 - sin(0.55)×cos(0.55)) - (0 - sin(0)×cos(0)) ≈ 0.0522

The shaded area, A_{1 shaded} = A₁ - A₂ = 0.075625 - 0.0522 ≈ 0.023425

Similarly, we have, between points 0.55 and 1.85

A₃ = 1/2 × (1.85 - 0.55) × 1/2 × (1.85 - 0.55) + (1.85 - 0.55) × 0.55/2 = 0.78

For y = sin²x, we have;

A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

8 0
2 years ago
Spraying the roof of a large building with cool water represents which type of cooling?
olganol [36]

Answer:

evaporative

Explanation:

The spraying water onto the ceiling can be considered as an evaporative cooling strategy.

<h2>Hope it is helpful....</h2>
4 0
2 years ago
This predator-prey graph tracks the wolf and moose populations in a certain ecosystem over the past hundred years. What is most
PSYCHO15rus [73]

Answer:

1,400

Explanation:

The carrying capacity is the maximum population size of the species that the environment can sustain indefinitely. When the population reaches its carrying capacity, it begins to decrease due to shortages of resources, such as food, water, habitat, and other necessities.

In the given example, we can see the moose population begins to grow when it reaches about 800 individuals. Then, the population keeps growing until it reaches above 1,400 individuals, after which it begins to decrease. Based on that, we can conclude that the carrying capacity of moose in this ecosystem is 1,400.

5 0
3 years ago
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