Answer: 38 is your answer hope this helped
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Step-by-step explanation:
Answer:
which agrees with option"B" of the possible answers listed
Step-by-step explanation:
Notice that in order to solve this problem (find angle JLF) , we need to find the value of the angle defined by JLG and subtract it from 
, since they are supplementary angles. So we focus on such, and start by drawing the radii that connects the center of the circle (point "O") to points G and H, in order to observe the central angles that are given to us as 
and 
. (see attached image)
We put our efforts into solving the right angle triangle denoted with green borders.
Notice as well, that the triangle JOH that is formed with the two radii and the segment that joins point J to point G, is an isosceles triangle, and therefore the two angles opposite to these equal radius sides, must be equal. We see that angle JOH can be calculated by : 
Therefore, the two equal acute angles in the triangle JOH should add to:
resulting then in each small acute angle of measure 
.
Now referring to the green sided right angle triangle we can find find angle JLG, using: 
Finally, the requested measure of angle JLF is obtained via: 
(x - 8)² = 144
Take square root of both sides
(x - 8) = √144
x - 8 = 12
x = 12 + 8
x = 20
Original lawn is = 20 feet * 20 feet square
QUESTION 1: In these type of question, the easiest way to get the answer is try to plug in the x and y values from the options given in the equation given, So in the first question all the choice except C are more then 14 if you plug in x and y's, for eg, if you plug in x = 3 and y = 2 , you get (3+3)2 = 14 6 x 2 = 14 12 is not equal to 14, so this eliminates this choice but if you chose C you get, (11+3)1 = 14 14 = 14 so this makes C the solution for first question and for the second question do the same thing, and the answer will be D. Hope this helps
QUESTION 2: 5xy + 9 = 44
5xy = 35
xy = 7
solution pairs are:
C. (1, 7) and (7, 1)
not mentioned: (-1.-7) and (-7, -1)
Hope this helps
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The area of the circle is 1260.25 π m^2
