The term of this sequence is:
<span>-(17/30)n^5+(113/12)n^4-(173/3)n^3+(1915/12)n^2-(5813/30)n+85 </span>
<span>Therefore,term number 7 is:-146/1=-146 </span>
<h3>
Answer: -20</h3>
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Work Shown:
Let x be the location of E on the number line.
Since C is the midpoint of E and F, this means we can find C's location by adding E and F together and dividing that sum by 2
midpoint = (endpoint1 + endpoint2)/2
C = (E+F)/2
Plug in E = x, C = -8 and F = 4. Then solve for x
C = (E+F)/2
-8 = (x+4)/2
(x+4)/2 = -8
x+4 = 2(-8) .... multiplying both sides by 2
x+4 = -16
x = -16-4 .... subtract 4 from both sides
x = -20
The location of point E on the number line is -20
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As a check, lets add E and F to get E+F = -20+4 = -16
Then cut this in half to get -16/2 = -8, which is the proper location of point C
This confirms our answer.
Answer:
The least number of stamps required is 
Step-by-step explanation:
Let the number of 
cent stamps be 
and 
cent stamps be 
We have
The minimum number is obtained when more cent stamps are used
Here cannot be greater than 
since 
Substitute 
Not possible since is not a fraction
Substitute 
Not possible since is not a fraction
Substitute 
Possible
Hence minimum number of stamps is
