A cylinder has a cross section of a circle as well as the sphere.
Hi! I am here! I need more letters lol but ya I am here
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 : The area of sector is, 
Step-by-step explanation :
Formula used to calculate the area of sector in degree is:
Area of sector = 
where,
= angle = 
r = radius = 25 mm
Now put all the given values in the above formula, we get:
Area of sector =
Area of sector = 
Area of sector =
Therefore, the area of sector is,
Answer:
C. A residual is the difference between the observed y-value of a data point and the predicted y-value on a regression line for the x-coordinate of the data point. A residual is positive when the point is above the line, negative when it is below the line, and zero when the observed y-value equals the predicted y-value.
Step-by-step explanation:
The residuals are obtained when there is some difference between the observed values and the fitted values of the data. Suppose we want to make a curve or hyperbola but the observed data does not actually give the curve required or there is some difference between the observed values and fitted values. The square of the sum of these differences is called residual.
The residual is positive when the point is above the line, negative when it is below the line, and zero when the observed y-value equals the predicted y-value.
Residual is obtained by subtracting the predicted value from observed value.This difference called the <u>residual</u> is
- positive when the observed value > predicted value
<em>For a positive value the point lies above the (fitted) line.</em>
- negative when the observed value < predicted value
<em>For a negative value the point lies below the (fitted) line.</em>
- zero when the observed value = predicted value
<em>For a zero value the point lies on the (fitted) line.</em>
Step-by-step explanation:
