Answer:
(1) F(n) = 8050 -400n
(2) 16 days
Step-by-step explanation:
Tylor first day he has left 7,250feet left to climb, this means that he has already climbed his 400ft on his first day
This will produce an arithmetic series for the distances left to climb as each day passes
initial climb = 7,250+400 = 7,650
Arithmetic Series
7650 , 7250 , 6850 , 6450 ......
a = 7650 , d = -400
(1) Explicit formula
F(n) = a + (n-1)d
= 7650 + (n-1)(-400)
= 7650 +400 -400n
F(n) = 8050 -400n
(2) 8050 -400n = 1750
6300 = 400n
n = 6300 /400 = 15.75 = 16 days( approximately)
So after 16days (approximately) Taylor will have 1750ft left to climb.
6 because it is the GCF. 36/6 is 6 and 30/6 is 5
For some reason I can’t see the picture sorry
F(x)=(3x^2+2x-5)/(x-4)
f(x)=(3x+5)(x-1)/(x-4)
x-4|(3x^2+2x-5)
(x-4)3x
3x^2+2x-5-(3x^2-12x)
(x-4)3x+14
14x-5-(14x-56)
51
Oblique=3x+14
Answer:
π/36 sec-1
Step-by-step explanation:
From the question we are given the formula below
s = rwt ..........eqn(1)
Where s= π/3m, r=3 m, t= 4 sec
We were not given value of " w" which implies we are required to find out values of" w"
We can make " w " the subject of the formula from eqn(1)
s = rwt
w= s/(rt)........eqn(2)
Then substitute for the values in eqn(2)
w= (π/3m ) / ( 3m × 4 sec)
= π/36 sec-1
Hence, the value of the
missing variable is π/36 sec-1
