Answer:
25π-24
Step-by-step explanation:
from the figure,
radius of the circle(r)=diameter of the circle(d)/2
=10/2
=5
Area of the circle(A)=πr^2
=π5^2
=25π
Again,
base of the right angle triangle(b)=6
perpendicular of right angle triangle(p)=8
we also know,
the area of right angle triangle = 1/2*base*height
=1/2*6*8
=1/2*48
=48/2
=24
Now, Area of the shaded region=Area of circle - Area of right angle triangle (given in the figure)
Area of shaded region= 25π -24
Answer:
To add or subtract functions, just add or subtract the values at each point where it makes sense. If the functions are given by formulas, you can just add or subtract the formulas (it doesn't matter whether you plug in values before or after).
Step-by-step explanation:
Answer:
Step-by-step explanation:
The point of this question is to find out the point where two lines intersect. First we need to get the equation of those lines
Slope of line 1:
(Yb -Ya)/(Xb - Xa) =
(-10 - (-14))/(-1 - (-3)) =
4/2 =
2
Use that slope to find the Y-intercept of line 1
y = 2x + b
-14 = 2(-3) +b
-14 = -6 + b
-8 = b
Therefore Line 1 is:
y = 2x - 8
Slope of line 2
(11 - 13)/(-1 - (-3)) =
-2/2 =
-1
Y-intercept of line 2
y = -x + b
13 = -(-3) +b
13 = 3 + b
10 = b
Therefore line 2 is
y = -x + 10
Now we have 2 equations to solve for the coordinates x and y
y = 2x - 8
y = -x + 10
Substitute y out in one of the equations
2x - 8 = -x + 10
3x = 18
x = 6
Plug x into one of the equations
y = 2(6) - 8
y = 12 - 8
y = 4
Therefore the solution is:
x=6, y=4
A.) For the Junior Varsity Team, mean would be the appropriate measure of center since the data is <span>symmetric or well-proportioned while we should use standard deviation for getting the measure of spread since it also measures the center and how far the values are from the mean.
b.) For the Varsity Team, the median would be the appropriate measure of the center since the data is skewed left and not evenly distributed so median could be used since it does not account for outliers while we use IQR or interquartile range in measuring the spread of data since IQR does not account for the data that is skewed. </span>