In each case use the new theorems to arrange the letters measures from greatest to least

Mathematics

1 answer:

tino4ka555 [31]2 years ago

7 0

Answer:

answer is a,b,c

Step-by-step explanation:

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Dr. Fitzgerald has graded 15 of 26 exams for Epi 501. (a) What proportion of all exams has Dr. Fitzgerald graded? (b) What was t

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Answer: a) 15:26, and b) 15:11.

Step-by-step explanation:

Since we have given that

Number of graded tests = 15

Number of total tests = 26

Number of ungraded tests is given by

$26-15\\\\=11$

a) Proportion of all exams has Dr. Fitxgerald graded is given by

15:26.

b) Ratio of graded to ungraded tests is given by 15:11

Hence, a) 15:26, and b) 15:11.

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4 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$