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Rus_ich [418]
3 years ago
8

A study participant was asked to observe circular images and determine if they were the same size as a 10 cm2 circle. circles wi

th areas greater than or equal to 11 cm2 were perceived as different. circles with areas between 10.1 and 10.9 cm2 were perceieved as the same size as the 10 cm2 circle. if the participant then was asked to compare samples with a circle with an area of 200 cm2, weber's law would predict that a circle of what area would be perceived as being of the same size?
Mathematics
1 answer:
Neko [114]3 years ago
6 0
<span>For the answer to the question above, according to Weber's law, the perceptible differences are proportional to the initial perception. So, with just a 10% increase in area was perceptible, but an increase of 1-9% in the area is considered imperceptible. The 9% increase in area reflected by 200 to 218 would be perceived as the same size as before.
The <u><em>answer is 218</em></u></span>
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Find a polynomial of degree 3 with real coefficients and zeros of -3,-1, and 4, for which f(-2)=-24
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We want to find a polynomial

<em>f(x)</em> = <em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em>

such that the roots of <em>f</em> are <em>x</em> = -3, <em>x</em> = -1, and <em>x</em> = 4, and <em>f(x)</em> takes on a value of -24 when <em>x</em> = -2.

The factor theorem for polynomials tells us that we can factorize <em>f(x)</em> as

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a</em> (<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4)

Expand the right side:

(<em>x</em> + 3) (<em>x</em> + 1) (<em>x</em> - 4) = <em>x</em>³ - 13<em>x</em> - 12

So we have

<em>a</em> <em>x</em>³ + <em>b</em> <em>x</em>² + <em>c</em> <em>x</em> + <em>d</em> = <em>a x</em>³ - 13<em>a</em> <em>x</em> - 12<em>a</em>

<em />

In order for both sides to be equal, the coefficients of both polynomials on terms of the same degree must be equal. This means

<em>a</em> = <em>a</em> (of course)

<em>b</em> = 0 (there is no <em>x</em>² term on the right)

<em>c</em> = -13<em>a</em>

<em>d</em> = -12<em>a</em>

<em />

We also have that <em>f</em> (-2) = -24, which means

<em>f</em> (-2) = <em>a</em> (-2 + 3) (-2 + 1) (-2 - 4)

-24 = 6<em>a</em>

<em>a</em> = -4

which in turn tells us that <em>c</em> = 52 and <em>d</em> = 48.

So we found

<em>f(x)</em> = -4<em>x</em>³ + 52<em>x</em> + 48

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3 years ago
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cluponka [151]
The answer is  A: x=1/2 , y= 1/3

Solve the following system:
{2 x + 3 y = 2 | (equation 1)
{6 x + 18 y = 9 | (equation 2)

Swap equation 1 with equation 2:
{6 x + 18 y = 9 | (equation 1)
{2 x + 3 y = 2 | (equation 2)

Subtract 1/3 × (equation 1) from equation 2:
{6 x + 18 y = 9 | (equation 1)
{0 x - 3 y = -1 | (equation 2)

Divide equation 1 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x - 3 y = -1 | (equation 2)

Multiply equation 2 by -1:
{2 x + 6 y = 3 | (equation 1)
{0 x+3 y = 1 | (equation 2)

Divide equation 2 by 3:
{2 x + 6 y = 3 | (equation 1)
{0 x+y = 1/3 | (equation 2)

Subtract 6 × (equation 2) from equation 1:
{2 x+0 y = 1 | (equation 1)
v0 x+y = 1/3 | (equation 2)

Divide equation 1 by 2:
{x+0 y = 1/2 | (equation 1)
v0 x+y = 1/3 | (equation 2)

Collect results:
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