Let O be the center of a circle. If <span>the measure of arc RS is 84 degrees, then m∠SOR=84^{0}. The triangle SOR is isoscales (because SO=OR as radii), so m∠RSO=m∠ORS=(180^{0}-84^{0}):2=48^{0}.
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Line RU is tangent to the circle in point R, this means that m∠ORU=90^{0}.
Consider the triangle SRU. m∠RSU=30^{0} and m∠SRU=48^{0}+90^{0}=138^{0}, then m∠RUS=180^{0}-30^{0}-138^{0}=12^{0}.
ANSWER: Correct choice B - 12^{0}.
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60a^5 is the answer to your question
1a) 8 / (1/2) = 16 * 3 = 48
1b) 3sqrt(49) = 3 * 7 = 21
1c) (5+2)(-8) / (-2)^3 -3
(7*-8) / (-8 -3)
-56/-11
56/11
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Let the area be y .
Area = (base) × (height)
Base = 2x
Height = h
Let the area of the rectangular pens be y .
∴ y = 2xh
Perimeter of all the fencing = 4x+3h
∴ 4x+3h = 120
now we solve for h
3h = 120-4x
h = 40 - 4/3 x
Now we will substitute this value in the above first equation:
y = 2xh
or, y = 2x (40 - 4/3 x)
or, y = 80x - 8/3 x²
Now for the maximum area we have to find the first order differentiation of y
now,
dy /dx = 80 - 16/3 x
At dy/dx = 0 we get the value of x for which y is maximum.
80 - 16/3 x = 0
or, - 16/3 x = -80
or, x = 15 feet
Hence height = 40 - 4/3 x = 40 - 20 = 20feet
Maximum area = 2xh = 2×15×40 = 1200 square feet
The dimensions of the rectangular pen should be 15 by 20 feet and the maximum area is 1200 square feet.
Disclaimer : The missing figure for the question is attached below.
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Answer:
x = 2.66666667
Step-by-step explanation:
3(x + 1) = 11
We multiply both x and 1 by 3 and get
3x + 3 = 11
Now we subtract 3 from both sides
3x + 3 - 3 = 11 - 3
3x = 8
Now divide both sides by 3
3x/3 = 8/3
x = 2.66666667
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