Answer:
89
Step-by-step explanation:
So the line segment CD is 12.7 and half that is 6.35. I wanted this 6.35 so I can look at the right triangle there and find the angle there near the center. This will only be half the answer. So I will need to double that to find the measure of arc CD.
Anyways looking at angle near center in the right triangle we have the opposite measurement, 6.35, given and the hypotenuse measurement, 9.06, given. So we will use sine.
sin(u)=6.35/9.06
u=arcsin(6.35/9.06)
u=44.5 degrees
u represented the angle inside that right triangle near the center.
So to get angle COD we have to double that which is 89 degrees.
So the arc measure of CD is 89.
Answer:
Step-by-step explanation:There are other kinds of numbers that can be graphed on the number line, too. ... which are natural numbers and which are whole numbers is to think of a “hole,” which can be represented by 0. ... Notice that distance is always positive or 0.
Answer:
60 degrees
Step-by-step explanation:
Restructured question:
The measure of two opposite interior angles of a triangle are x−14 and x+4. The exterior angle of the triangle measures 3x-45 . Solve for the measure of the exterior angle.
First you must know that the sum of interior angle of a triangle is equal to the exterior angle
Interior angles = x−14 and x+4
Sum of interior angles = x-14 + x + 4
Sum of interior angles = 2x - 10
Exterior angle = 3x - 45
Equating both:
2x - 10 = 3x - 45
Collect like terms;
2x - 3x = -45 + 10
-x = -35
x = 35
Get the exterior angle:
Exterior angle = 3x - 45
Exterior angle = 3(35) - 45
Exterior angle = 105 - 45
Exterior angle = 60
Hence the measure of the exterior angle is 60 degrees
<em>Note that the functions of the interior and exterior angles are assumed. Same calculation can be employed for any function given</em>
Answer:
x= - 1.48 or 0.48
Step-by-step explanation:
I used general formula. Hope I'm on a right path.
Answer:
The wedge cut from the first octant ⟹ z ≥ 0 and y ≥ 0 ⟹ 12−3y^2 ≥ 0 ⟹ 0 ≤ y ≤ 2
0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2
V = ∫∫∫ dzdydx
dz has changed from zero to 12−3y^2
dy has changed from zero to 2-x
dx has changed from zero to 2
V = ∫∫∫ dzdydx = ∫∫ (12−3y^2) dydx = ∫ 12(2-x)-(2-x)^3 dx =
24(2)-6(2)^2+(2-2)^4/4 -(2-0)^4/4 = 20
Step-by-step explanation:
