3. f(-6) = 12+1 =13
f(-2) = 4+1 = 5
f(0) =1
Range {1,5,13}
4. f(-2) = (-2)^3+1 =-7
f(-1) = (-1)^2 +1 =0
f(3) = (3)^3 +1 = 28
Range = {-7,0,28}
5.the sequence is arithmetic
d= -11+19 = 8
an = a1 + d(n-1)
an = -19 +8(n-1)
6.l =w+5
a =l*w
a(w) =(w+5) * w
a(w)= w^2 +5w
f(w) = w^2 +5w
f(8) = 8^2 +5(8)
f(8) = 64 +40
f(8) =104 in^2
Answer:Input : A = 12, B = 8
Output : X = 2, Y = 10
Input : A = 12, B = 9
Output : -1
Step-by-step explanation:
Answer: 29.28 years
Explanation:
From Kepler's third law the square of orbital period of revolving celestial body is proportional to the cube of semi -major axis from the body it is revolving about.
P² =A³
Where, P is the orbital period in years and A is the semi-major axis in AU (Astronomical units)
It is given that, For Saturn, A = 9.5 AU. We need to find P
⇒P² = (9.5 AU)³
⇒P² = 857.38
⇒P = 29.28 years
Thus, the orbital period of Jupiter is 29.28 years around the Sun.
(-2/3) (-1.6)
(−2)(−3602879701896397) / (3)(2251799813685248)
=7205759403792794 / 6755399441055744
=1.066667
