The picture of the problem has been attached below :
Answer:
13.5
Step-by-step explanation:
Applying the sine rule to solve for x
SinA /a = SinB / b = SinC/ c
Sin 34 / x = Sin 27/11
Cross multiply :
11 * sin34 = x * sin 27
6.1511219 = 0.4539904x
Divide both sides by 0.4539904
6.1511219/0.4539904 = x
13.549 = x
x = 13.5
Answer:
There is a similar problem with the same numbers except with no negatives so it goes 1 , 3 , 9 , 27, 81 for this question the answer is that it diverges.
Step-by-step explanation:
It was on the unit test review for edge and the answer was that it diverges.
Answer:
-(x+2)/(x+8)
Step-by-step explanation:
( x^2 -x-6)
------------------
24 - 5x -x^2
Factor out a minus sign from the denominator
( x^2 -x-6)
------------------
-( x^2 +5x -24)
Factor the numerators and the denominators
( x-3) (x+2)
------------------
-( x+8)(x-3)
Cancel like terms
x+2
------------------
-(x+8)
Answer:
Step-by-step explanation:
Find c if ∠PQR = 90°?
I will ASSUME you mean point P is at (3. c)
slope of PQ is (2 - c) / (9 - 3) = (2 - c) / 6
slope of QR is (11 - 2) / (3c - 9) = 9 / (3c - 9)
perpendicular lines have negative reciprocal slopes.
(2 - c) / 6 = -1(3c - 9)/9
9(2 - c) = -6(3c - 9)
18 - 9c = -18c + 54
9c = 36
c = 4
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,
so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
