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Reason for why limits not existing : In the previous section we saw two limits that did not.
We saw that
did not exist since the function did not settle down to a single value as t approached t = 0 . The closer to t = 0 we moved the more passionately the function oscillated & in order for a limit to exist the function have to settle down to a single value.
However we saw that did not present not since the function didn't settle down to a single number as we moved in towards t = 0 , but rather then because it settled into two distinct numbers based on which side of t = 0 we were on.
The problem was that, as we approached t =0 , the function was moving in towards different numbers on each of the side.
If the p th term of an AP is q and the q th term is p. P.T its n th term is (p+q-n). Ans: APQ a p = q a q = p a n = ? a + (p-1) d = q a + (q-1) d = p
what is -6.4 as a fraction?
The production costs per week for generating x widgets is given by, C ( x ) = 500 + 350 x - 0.09 x 2 , 0 ≤ x ≤ 1000 Answer following questions. (a) What is the c
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how do i multiply and divide fractions?
Ask question #Min 4.4238/[1.047+{1.111*[9.261/7.777]}*1.01
Last year Jonathan was 603/4 inches tall. This year he is 651/4 inches tall. How many inches did he grow? Subtract to find outthe difference in heights. You will need to borro
Evaluate the subsequent integral. ∫ (tan x/sec 4 x / sec 4 x) dx Solution This kind of integral approximately falls into the form given in 3c. It is a quotient of ta
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Ok this is true or false wit a definition. The GCF of a pair of numbers can never be equal to one of the numbers.
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