Answer:
1) The 38th term is A. 38
2) The 11th term is D. 3072
3) The 3rd term is C. 34
Step-by-step explanation:
1) I added 2 until I got to the 38th term.
2) I multiplied until I got to the 11th term.
3) I added 3 until i got to the 12th term.
The teacher showed a more complex way to do it but this is just what I did.
I took the quiz, so I know that I got them all correct.
Answer:
Step-by-step explanation:
To calculate the weighted moving average for period 13 with weights 0.4 and 0.3.
P13 = (30.7x 0.4) + (42.0x 0.3)
P13 = 12.28 + 12.60
P13 = 24.88
Answer:

Step-by-step explanation:
we know that
The area of the shaded region is equal to the area of the circle minus the area of the two isosceles triangles
so
Find the area of the circle
The area of the circle is equal to

where r is the radius of the circle
In this problem we have

substitute


Find the area of the two triangles
The area of the two isosceles triangles is equal to
![A=2[\frac{1}{2}bh]=bh](https://tex.z-dn.net/?f=A%3D2%5B%5Cfrac%7B1%7D%7B2%7Dbh%5D%3Dbh)
we have


substitute

Find the area of the shaded region

Answer:
m∠FJH=60°
Step-by-step explanation:
The complete question is
JG bisects FJH, FJG= (2x + 4)° and GJH = (3x -9)°
What is FJH
we know that
m∠FJH=m∠FJG+m∠GJH -----> equation A
If ray JG is an angle bisector of ∠FJH
then
m∠FJG=m∠GJH -----> equation B
substitute the given values in equation B and solve for x
(2x + 4)°=(3x -9)°
3x-2x=4+9
x=13
Find the measure of angle FJH
m∠FJH=(2x + 4)°+(3x -9)°
substitute the value of x
m∠FJH=(2(13) + 4)°+(3(13) -9)°
m∠FJH=(30)°+(30)°
m∠FJH=60°