<span>(6y3 + 17y − 3) − (4y3 − 11y + 9)
</span>= 6y3 + 17y − 3 − 4y3 + 11y - 9
= 2y3 + 28y - 12
Find, correct to the nearest degree, the three angles of the triangle with the vertices d(0,1,1), e( 2, 4,3) − , and f(1, 2, 1)
Ksju [112]
Well, here's one way to do it at least...
<span>For reference, let 'a' be the side opposite A (segment BC), 'b' be the side opposite B (segment AC) and 'c' be the side opposite C (segment AB). </span>
<span>Let P=(4,0) be the projection of B onto the x-axis. </span>
<span>Let Q=(-3,0) be the projection of C onto the x-axis. </span>
<span>Look at the angle QAC. It has tangent = 5/4 (do you see why?), so angle A is atan(5/4). </span>
<span>Likewise, angle PAB has tangent = 6/3 = 2, so angle PAB is atan(2). </span>
<span>Angle A, then, is 180 - atan(5/4) - atan(2) = 65.225. One down, two to go. </span>
<span>||b|| = sqrt(41) (use Pythagorian Theorum on triangle AQC) </span>
<span>||c|| = sqrt(45) (use Pythagorian Theorum on triangle APB) </span>
<span>Using the Law of Cosines... </span>
<span>||a||^2 = ||b||^2 + ||c||^2 - 2(||b||)(||c||)cos(A) </span>
<span>||a||^2 = 41 + 45 - 2(sqrt(41))(sqrt(45))(.4191) </span>
<span>||a||^2 = 86 - 36 </span>
<span>||a||^2 = 50 </span>
<span>||a|| = sqrt(50) </span>
<span>Now apply the Law of Sines to find the other two angles. </span>
<span>||b|| / sin(B) = ||a|| / sin(A) </span>
<span>sqrt(41) / sin(B) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(41) / sqrt(50) = sin(B) </span>
<span>.8222 = sin(B) </span>
<span>asin(.8222) = B </span>
<span>55.305 = B </span>
<span>Two down, one to go... </span>
<span>||c|| / sin(C) = ||a|| / sin(A) </span>
<span>sqrt(45) / sin(C) = sqrt(50) / .9080 </span>
<span>(.9080)sqrt(45) / sqrt(50) = sin(C) </span>
<span>.8614 = sin(C) </span>
<span>asin(.8614) = C </span>
<span>59.470 = C </span>
<span>So your three angles are: </span>
<span>A = 65.225 </span>
<span>B = 55.305 </span>
<span>C = 59.470 </span>
Answer:
pink's area: 114
teal's area: 183
Step-by-step explanation:
lmk if you need the work
For this, you would need to find 70% of 140. To do so you must do the following: 140 x 0.7, which would lead you to the following answer: 98. So, if someone were to weigh 140 pounds, they would have around 98 pounds of water in their body!
I hope this was helpful, and if you need any more help, please let me know so I can further assist you.
