Check the picture below to the left, let's use those sides with the law of sines
![\textit{Law of sines} \\\\ \cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{sin(14^o)}{97}=\cfrac{sin(84^o)}{XZ}\implies XZ = \cfrac{97\cdot sin(84^o)}{sin(14^o)}\implies XZ \approx 398.76 \\\\\\ \stackrel{\textit{now using SOH CAH TOA}}{cos(82^o) = \cfrac{XW}{XZ}}\implies XZcos(82^o)=XW \\\\\\ 398.76cos(82^o)\approx XW\implies 55.497\approx XW\implies \stackrel{\textit{rounded up}}{55=XW}](https://tex.z-dn.net/?f=%5Ctextit%7BLaw%20of%20sines%7D%20%5C%5C%5C%5C%20%5Ccfrac%7Bsin%28%5Cmeasuredangle%20A%29%7D%7Ba%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20B%29%7D%7Bb%7D%3D%5Ccfrac%7Bsin%28%5Cmeasuredangle%20C%29%7D%7Bc%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7Bsin%2814%5Eo%29%7D%7B97%7D%3D%5Ccfrac%7Bsin%2884%5Eo%29%7D%7BXZ%7D%5Cimplies%20XZ%20%3D%20%5Ccfrac%7B97%5Ccdot%20sin%2884%5Eo%29%7D%7Bsin%2814%5Eo%29%7D%5Cimplies%20XZ%20%5Capprox%20398.76%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bnow%20using%20SOH%20CAH%20TOA%7D%7D%7Bcos%2882%5Eo%29%20%3D%20%5Ccfrac%7BXW%7D%7BXZ%7D%7D%5Cimplies%20XZcos%2882%5Eo%29%3DXW%20%5C%5C%5C%5C%5C%5C%20398.76cos%2882%5Eo%29%5Capprox%20XW%5Cimplies%2055.497%5Capprox%20XW%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%7D%7D%7B55%3DXW%7D)
<h3>Given</h3>
tan(x)²·sin(x) = tan(x)²
<h3>Find</h3>
x on the interval [0, 2π)
<h3>Solution</h3>
Subtract the right side and factor. Then make use of the zero-product rule.
... tan(x)²·sin(x) -tan(x)² = 0
... tan(x)²·(sin(x) -1) = 0
This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:
... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)
Then our equation becomes
... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0
... -sin(x)²/(1 +sin(x)) = 0
Now, we know the only solutions are found where sin(x) = 0, at ...
... x ∈ {0, π}
Answer:
[A] It will take 24 minutes for Joe and Charmaine to meet again at the starting line.
[B] Joe will complete 2 laps
[C] Charmaine will complete 3 laps.
Step-by-step explanation:
Factor two numbers 12 and 8:
12 = 2×2×3
8 = 2×2×2
Find LCM of 12 and 8:
Lcm of 12 and 8 - 2×2×3×2 = 24
Therefore,
[A] How much time does it take for them to meet again at the starting line?
It will take 24 minutes for Joe and Charmaine to meet again at the starting line.
[B] How many laps does Joe complete
Joe will complete 2 laps
[C] How many laps does Charmaine complete?
Charmaine will complete 3 laps.
<u>~lenvy~</u>
Answer:
-6 I meant
Step-by-step explanation:
Y - y1 = m(x - x1)
slope(m) = -4/3
(0,-12)....x1 = 0 and y1 = -12
sub
y - (-12) = -4/3(x - 0) =
y + 12 = -4/3(x - 0) <=== point slope form
y + 12 = -4/3x
y = -4/3x - 12 <=== slope intercept form
y = -4/3x - 12
4/3x + y = -12
4x + 3y = -36 <=== standard form