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Leviafan [203]
3 years ago
5

Triangle JKL is equilateral. One side of the triangle, JL, is a diameter of circle M. Which is true about line segments JK and K

L?
Both segments are tangent to circle M but are not chords.
One segment is tangent to circle M and one segment is a chord in circle M.
Both segments are chords in circle M but are not tangents.
Neither segment is a chord nor tangent to circle M.
Mathematics
2 answers:
antoniya [11.8K]3 years ago
7 0
It could be concluded that since line JL is a diameter of a circle, therefore, JK and KL are the tangents of the specific circle that JL is part of. By definition, a tangent is "a line that touches a curve at one point without touching it." Both these lines touches the edges of the circle M therefore they are tangents.
Agata [3.3K]3 years ago
4 0

Answer:

Just took the test, it's D!


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A repeated-measures study comparing two treatments with n = 4 participants produces md = 2 and ss = 75 for the difference scores
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The estimated standard error for the sample mean difference is  2.5 .

According to the question

A repeated-measures study comparing two treatments

n = 4

MD(mean difference) = 2

SS (sum of square) = 75

Now,

error for the sample  

Formula for standard error

S^{2} = \frac{SS}{n-1}

by substituting the value

S^{2} = \frac{75}{4-1}

S^{2} = \frac{75}{3}

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Standard error of the estimate for the sample mean difference

As

The standard error of the estimate is the estimation of the accuracy of any predictions.

The formula for standard error of the mean difference

standard error of the mean difference  =\frac{standard\\\ error}{\sqrt{n} }  

standard error of the mean difference = \frac{5}{\sqrt{4} }  

standard error of the mean difference = 2.5

Hence, the estimated standard error for the sample mean difference is  2.5 .

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<u>Answer-</u>

<em>The polynomial function is,</em>

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<u>Solution-</u>

The zeros of the polynomial are 2 and (3+i). Root 2 has multiplicity of 2 and (3+i) has multiplicity of 1

The general form of the equation will be,

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\Rightarrow y=(x-2)^2(x-3-i)(x-3+i)

\Rightarrow y=(x^2-4x+4)((x-3)-i)((x-3)+i)

\Rightarrow y=(x^2-4x+4)((x-3)^2-i^2)

\Rightarrow y=(x^2-4x+4)((x^2-6x+9)+1)

\Rightarrow y=(x^2-4x+4)(x^2-6x+10)

\Rightarrow y=x^2x^2-6x^2x+10x^2-4x^2x+4\cdot \:6xx-4\cdot \:10x+4x^2-4\cdot \:6x+4\cdot \:10

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Therefore, this is the required polynomial function.


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