Answer:
O is the center of the circle with radius IE(=ID=EF)
Step-by-step explanation:
Join all 3 points D, E, F, forming the triangle DEF.
Let the midpoint of EF be M and the midpoint of ED be N. (first picture)
Join point I to E, D and F.
Since IN is both an altitude and median to triangle EID, then triangle EID is an isosceles triangle, and IE=ID
similarly, we see that IE=IF.
conclusion: IE=ID=EF.
Answer:heres what i got
The left side −0
¯4 is greater than the right side−0.¯¯¯¯¯¯6h means that the given statement is always true.
Step-by-step explanation:
Volume = (PI * r^2 * h) / 3
r^2 = (3 * Volume) / (PI * h)
r = Square Root (3 * Volume) / (PI * h)
Miguel: 500 out of 750 students have part time jobs.
500 ÷ 250 = 2
750 ÷ 250 = 3
500:750 = 2:3
A) 200 out of 300 ⇒ 200/100 and 300/100 ⇒ 2:3
B) 700 out of 1100 ⇒ 700/100 and 1100/100 ⇒ 7:11
C) 800 out of 1200 ⇒ 800/400 and 1200/400 ⇒ 2:3
D) 9000 out of 1300 ⇒ 9000/100 and 1300/100 ⇒ 90:13
Among the choices, Choice B could represent Kureshi's Data because it is not proportional to the data of Miguel.
Choice D is not possible. You cannot have a result that is way beyond the scope of your population. It is impossible to get 9000 students out of only 1300 students.
csc(2x) = csc(x)/(2cos(x))
1/(sin(2x)) = csc(x)/(2cos(x))
1/(2*sin(x)*cos(x)) = csc(x)/(2cos(x))
(1/sin(x))*1/(2*cos(x)) = csc(x)/(2cos(x))
csc(x)*1/(2*cos(x)) = csc(x)/(2cos(x))
csc(x)/(2*cos(x)) = csc(x)/(2cos(x))
The identity is confirmed. Notice how I only altered the left hand side (LHS) keeping the right hand side (RHS) the same each time.
