At a local park there is a large compass painted on the sidewalk. Blaine starts at due north and walks 150° in a clockwise direc
tion. Using the unit circle, find the sine value of his current position.
A) negative square root of 3 over 2.
B) square root 2 over 2.
C) one half.
D) negative one half.
A) negative square root of 3 over 2.
B) square root 2 over 2.
C) one half.
D) negative one half.
1 answer:
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Answer:
1.B
2.A
3. B
Step-by-step explanation:
1.
We have the denominator of the fraction as following:
As the initial one is a fraction, so that its denominator has to be different from 0.
=> () ≠ 0
⇔ (x +1) (x +5) ≠ 0
⇔ (x + 1) ≠ 0; (x +5) ≠ 0
⇔ x ≠ -1; x ≠ -5
Replace it into the initial equation, we have:
As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)
=>
So that with x ≠ 1; x ≠ -5
So that the answer is B.
2.
As the initial one is a fraction, so that its denominator has to be different from 0
=> x + 4 ≠ 0
=> x ≠ -4
As is also a fraction, so that its denominator (x-1) has to be different from 0
=> x - 1 ≠ 0
=> x ≠ 1
We have an equation:
=>
Replace it into the initial equation, we have:
As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)
=>
So that the initial equation is equal to with x ≠-4; x ≠1
=> So that the correct answer is A
3.
As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0
We have:
(4x + x^2) = 4x + x.x = x ( x + 4)
So that: (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0
⇔ ⇔
As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:
As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:
So that with x ≠ 0; x ≠ -4
=> The correct answer is B
Answer:
720 ways
Step-by-step explanation:
Generally, combination is expressed as;
The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.
⇒ ⁹C₇
= 36
The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.
⇒ ⁶C₃
= 20
Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.
⁹C₇ × ⁶C₃
= 36 × 20
= 720 ways