Two types of monocular depth cues are
A. motion parallax.. pictorial depth cues
B. convergence...retinal disparity
C. motion parallax retinal disparity
D. convergence.
pictorial depth cues

Answer:

.03

Explination

I am god

Answer:

There are many signs of having social anxiety. You might have some signs/symptoms but not the diagnosis.

Explanation:

Signs that you might have social anxiety:

Fear of situations in which you may be judged

Worrying about embarrassing or humiliating yourself

Intense fear of interacting or talking with strangers

Fear that others will notice that you look anxious

Fear of physical symptoms that may cause you embarrassment, such as blushing, sweating, trembling or having a shaky voice

Avoiding doing things or speaking to people out of fear of embarrassment

Avoiding situations where you might be the center of attention

Having anxiety in anticipation of a feared activity or event

Enduring a social situation with intense fear or anxiety

Spending time after a social situation analyzing your performance and identifying flaws in your interactions

Expecting the worst possible consequences from a negative experience during a social situation

Physical Signs:

Blushing

Fast heartbeat

Trembling

Sweating

Upset stomach or nausea

Trouble catching your breath

Dizziness or lightheadedness

Feeling that your mind has gone blank

Muscle tension

Things that could help:

Interacting with unfamiliar people or strangers

Attending parties or social gatherings

Going to work or school

Starting conversations

Making eye contact

Dating

Entering a room in which people are already seated

Returning items to a store

Eating in front of others

Using a public restroom

For a particle with some position function y(t), we're given its velocity function v(t) = 1 - tan⁻¹(eᵗ ) and we're told that it starts with a position of y = -1 at the start.

(a) Acceleration is the rate of change of velocity, so to get the acceleration function a(t), you have to differentiate v(t) :

a(t) = d/dt v(t)

a(t) = d/dt [1 - tan⁻¹(eᵗ )]

a(t) = - d/dt [tan⁻¹(eᵗ )]

Apply the chain rule: take f(u) = tan⁻¹(u) and u(t) = eᵗ, so that

df/dt = df/du × du/dt

→   a(t) = - 1/(1 + u ²) × eᵗ = - eᵗ / (1 + e²ᵗ )

Then when t = 2, the acceleration is

a (2) = - e² / (1 + e⁴) ≈ -0.1329

(b) The acceleration of the particle at t = 2 is negative, so the particle's speed at this moment is decreasing. (Recall the first derivative test.)

(c) Note that at t = 0, the particle's velocity is v (0) = 1 - tan⁻¹(e⁰) = 1 - π/4 ≈ 0.2416 > 0. This means that the particle is moving upward at the start.

From part (a), we see that the acceleration is always negative, so the particle would be slowing down until it reaches zero velocity, after which point its velocity will become negative (meaning its speed would increase again as it starts to accelerate in the negative direction).

This means the particle's highest position on the y-axis occurs at the same time v(t) becomes 0, so you solve:

1 - tan⁻¹(eᵗ ) = 0

1 = tan⁻¹(eᵗ )

tan(1) = tan(tan⁻¹(eᵗ ))

tan(1) = eᵗ

ln(tan(1)) = ln(eᵗ )

ln(tan(1)) = t ≈ 0.443

(d) Use the fundamental theorem of calculus to find an expression for y(t) :

y(t) = y (0) + ∫₀ᵗ v(s) ds

(where s is just a dummy variable)

Then the particle's position at t = 2 is

y (2) = y (0) + ∫₀² v(s) ds

v(t) unfortunately does not have a nice antiderivative, so you'll have to use a calculator. (I hope this question is given in the calculator portion!) You would find

∫₀² v(s) ds ≈ -0.3607

so that

y (2) ≈ -1 + (-0.3607) ≈ -1.3607

The sign of the particle's velocity at this time tells you in which way the particle is moving. We have

v (2) = 1 - tan⁻¹(e²) ≈ -0.4363

which is negative, so the particle is moving away from the origin. This is consistent with the answer found in part (c):

• before t ≈ 0.443, the particle is moving upward toward the origin;

• at t ≈ 0.443, the particle reaches its highest point at y (2) ≈ -1.3607, then turns around and accelerates back downward and away from the origin

Answer:

as a korean, living in america and going to schools with many white kids was tough. during the virus, i have experienced many moments where i have been stared down with scared faces, or been cussed out, telling me to go back to where i came from. even in school, the kids would do the normal things, and stretch out their eyes to mimic mine, which actually are pretty big. teachers ignored me more or asked insensitive questions, and all my life, i have felt a barrier, and social inequality. now that the korean things are trending though, many know who i am, but it is annoying for them to connect me to korean beauty, korean plastic surgery, kpop, and other things that managed to become popular in america.  

Explanation:

All of the following except labels are possible side effects of stigmas associated with mental disorders. Thus, option B is correct.

Social stigma refer to the discrimination or the disapproval against the individual or the group that should be based on the social characteristics where it should be make a difference from the other society members.

Its side effects include the discrimination, prejudice. negative stereotypes, etc.The possible side effects of stigmas associated with mental disorders should not include the labels.

In social science and anthropology, as well as in everyday English discourse, the term "ethnocentrism" refers to the practise of using one's own culture or ethnicity as a frame of reference to evaluate the Other cultures.

Learn more about stigma on:

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