Answer/Step-by-step explanation (ac > b² or b² < ac.
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A/c to question, we have to show:-
b² >ac in A.P ........ (1)
b² = ac in G.P .....(2)
b² < ac in H.P. ..... (3)
b = a+c/2 (A.P)
b = √ac ( G.P)
b = 2ac/a+c (H.P)
In A.P :
b² > ac = b² - ac
= (a+c/2)² - ac
= (a²+2ac+c²/4) - ac = a² + 2ac + c² - 4ac / 4
= a² - 2ac + c² / 4 = ( a - c ) ² / 4 > 0 Hence, b²>ac
In G.P:-
b = √ac
Hence, b² = ac
In H.P :- b² < ac = ac > b² = ac - b² = ac - ( 2ac / a+c)
= ac(a+c) - 2ac / a+c
= a²c + ac² - 2ac / a+c
= ac(2ac - 2) / a+c > 0
Hence, ac > b² or b² < ac.
Answer:
They each have 64
Step-by-step explanation:
let C = number of fish that Cecilia has --- then, O = 2C, where O = number of fish that Oscar has --- and T = O + 8 = 2C + 8, where T = number of fish that Tasha has --- then, C + O + T = C + 2C + (2C + 8) = 148 --- or, 5C + 8 = 148 --- or, 5C = 140 --- or, C = 140/5 = 28 --- and, O = 2C = 2(28) = 56 --- and, T = O + 8 = 56 + 8 = 64
Well there similar because your adding 10 more or 10× since from ones to the tens is 10×
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
Answer:
It's rational
Step-by-step explanation:
It's an integer and all integers are rational
