Answer:
2,5
Step-by-step explanation:
Answer:
Step-by-step explanation:
a1=2/3
sequence is 2/3,3/4,4/5,...
for numerator a1=2
d=3-2=1
numerator of nth term=a1+(n-1)d=2+(n-1)×1=2+n-1=n+1
denominator = 1 more than numerator=n+1+1=n+2
so an=(n+1)/(n+2)
or for denominator a1=3,d=4-3=1
denominator of nth term=3+(n-1)×1=3+n-1=n+2
an=(n+1)/(n+2)
9514 1404 393
Answer:
4) 6x
5) 2x +3
Step-by-step explanation:
We can work both these problems at once by finding an applicable rule.
where O(h²) is the series of terms involving h² and higher powers. When divided by h, each term has h as a multiplier, so the series sums to zero when h approaches zero. Of course, if n < 2, there are no O(h²) terms in the expansion, so that can be ignored.
This can be referred to as the <em>power rule</em>.
Note that for the quadratic f(x) = ax^2 +bx +c, the limit of the sum is the sum of the limits, so this applies to the terms individually:
lim[h→0](f(x+h)-f(x))/h = 2ax +b
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4. The gradient of 3x^2 is 3(2)x^(2-1) = 6x.
5. The gradient of x^2 +3x +1 is 2x +3.
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If you need to "show work" for these problems individually, use the appropriate values for 'a' and 'n' in the above derivation of the power rule.
Answer:yes she does cauce she has 1.05
Step-by-step explanation:
That’s really rude man, be nice to people