Answer: 272
Step-by-step explanation:
Let a1=11
a2=20
a3=29
Formula for sequence=
an=a1+(n-1)d
an = nth term
a1= first term
n= nth position
d= common difference
We are looking for the 30th term,so our n=30
d= a2-a1
d= 20-11
d= 9
Using the formula
an= a1+(n-1)d
a30= 11+(30-1)9
a30= 11+(29)9
a30= 11+(29×9)
a30= 11+261
a30= 272
Therefore, the 30th term is 272
Answer:

Step-by-step explanation:
step 1:- by using partial fractions
......(1)
<u>step 2:-</u>
solving on both sides

substitute x =0 value in equation (2)
1=A(1)+0
<u>A=1</u>
comparing x^2 co-efficient on both sides (in equation 2)
0 = A+B
0 = 1+B
B=-1
comparing x co-efficient on both sides (in equation 2)
<u>-</u>1 = C
<u>step 3:-</u>
substitute A,B,C values in equation (1)
now

by using integration formulas
i) by using
.....(b)
.....(c)
<u>step 4:-</u>
by using above integration formulas (a,b,and c)
we get answer is

It would be: 420 + (420 * 17.5%)
= 420 + (420 * 0.175) [ 17.5% = 0.175 ]
= 420 + 73.5
= 493.5
In short, Your Answer would be <span>£493.5
Hope this helps!</span>
Answer: 0
Explanation:
-1/2n -(-1/2n)
= -1/2n + 1/2n
= 0
a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as
... c = k·s² + m·s + n
Filling in the given values gives three equations in k, m, and n.

Subtracting each equation from the one after gives

Subtracting the first of these equations from the second gives

Using the next previous equation, we can find m.

Then from the first equation
[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]
There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with
... c = 0.01s² - s + 37
b) At 150 kph, the cost is predicted to be
... c = 0.01·150² -150 +37 = 112 . . . cents/km
c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.
d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.