A) Tangent is positive in Quadrant I.
Since sine and cosine are both positive in Quadrant I and tangent is the ratio of sine to cosine, tangent is positive in Quadrant I
2/33 * 1/5 * 11/10 = (2 * 1 * 11) / (33 * 5 * 10) = 22/1650 reduces to 1/75
Answer:
x = 3
x = (-1)/2
x = 13/4
Step-by-step explanation:
Solve for x:
(2 x)/3 + 15 = 17
Put each term in (2 x)/3 + 15 over the common denominator 3: (2 x)/3 + 15 = (2 x)/3 + 45/3:
(2 x)/3 + 45/3 = 17
(2 x)/3 + 45/3 = (2 x + 45)/3:
1/3 (2 x + 45) = 17
Multiply both sides of (2 x + 45)/3 = 17 by 3:
(3 (2 x + 45))/3 = 3×17
(3 (2 x + 45))/3 = 3/3×(2 x + 45) = 2 x + 45:
2 x + 45 = 3×17
3×17 = 51:
2 x + 45 = 51
Subtract 45 from both sides:
2 x + (45 - 45) = 51 - 45
45 - 45 = 0:
2 x = 51 - 45
51 - 45 = 6:
2 x = 6
Divide both sides of 2 x = 6 by 2:
(2 x)/2 = 6/2
2/2 = 1:
x = 6/2
The gcd of 6 and 2 is 2, so 6/2 = (2×3)/(2×1) = 2/2×3 = 3:
Answer: x = 3
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Solve for x:
3 x - x + 8 = 7
Grouping like terms, 3 x - x + 8 = (3 x - x) + 8:
(3 x - x) + 8 = 7
3 x - x = 2 x:
2 x + 8 = 7
Subtract 8 from both sides:
2 x + (8 - 8) = 7 - 8
8 - 8 = 0:
2 x = 7 - 8
7 - 8 = -1:
2 x = -1
Divide both sides of 2 x = -1 by 2:
(2 x)/2 = (-1)/2
2/2 = 1:
Answer: x = (-1)/2
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Solve for x:
4 (2 x - 6) = 2
Divide both sides of 4 (2 x - 6) = 2 by 4:
(4 (2 x - 6))/4 = 2/4
4/4 = 1:
2 x - 6 = 2/4
The gcd of 2 and 4 is 2, so 2/4 = (2×1)/(2×2) = 2/2×1/2 = 1/2:
2 x - 6 = 1/2
Add 6 to both sides:
2 x + (6 - 6) = 1/2 + 6
6 - 6 = 0:
2 x = 1/2 + 6
Put 1/2 + 6 over the common denominator 2. 1/2 + 6 = 1/2 + (2×6)/2:
2 x = 1/2 + (2×6)/2
2×6 = 12:
2 x = 1/2 + 12/2
1/2 + 12/2 = (1 + 12)/2:
2 x = (1 + 12)/2
1 + 12 = 13:
2 x = 13/2
Divide both sides by 2:
x = (13/2)/2
2×2 = 4:
Answer: x = 13/4
