Recent content by ProBasket

let's say i had 115% for my online homework which is worth 16% of the final grade and 4% for my written homework which is worth 8% of the final grade. how would i find my total percent for both online+hw together? I am thinking (115/2 + 115 + 2)/2 = 87.25% for both written homework and online...

a car starts from rest and acclerates down a straight track of length L= 1600m with a constant acceleration. If the time it takes for the car to travel the final d= 100 m of the track (from 1500m to 1600m) is T=0.125s, then the acceleration of the car is... the answer is 80.7 m/s^2 this isn't...

i don't think i modeled it incorrectly, because there's a similar problem in the book, but maybe i made a mistake so who knows, but here are the first few terms... 10 - 1000x^2 + 10000x^4 -10000000x^6 + 1000000000x^8... coefficients of the odd powers are zero... but i can't seem to get...

the function f(x) = \frac{10}{1+100*x^2} is represented as a power series f(x) = \sum_{n=0}^{\infty} C_nX^n Find the first few coefficients in the power series: C_0 = ____ C_1 = ____ C_2 = ____ C_3 = ____ C_4 = ____ well f(x) = \frac{10}{1+100*x^2} can be written as...

well the 4th term is \frac{(0.3)^6}/{6!} but it comes out to .000001 and the 5th term has 8 zeros, so the 4th term is closer to the value 0.0000001. so is that how the book got 4th term as an answer?

\sum_{n=1}^{\infty} a_n = 1 - \frac {(0.3)^2}{2!} + \frac {(0.3)^4}{4!} - \frac {(0.3)^6}{6!} + \frac {(0.3)^8}{8!} - ... how many terms do you have to go for your approximation (your partial sum) to be within 0.0000001 from the convergent value of that series? the answer to this...

so A is false because there is a 2 in front of the 'sum of' sign(which you can't do)? and B and C are true?

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is...

Test each of the following series for convergence by either the Comparison Test or the Limit Comparison Test. \sum_{n=1}^\infty \frac{2n^4}{n^5+7} this diverges using the p-series and comparison test right? p <1 \sum_{n=1}^\infty \frac{2n^4}{n^9+7} and this converges right? because p >...

i forgot the .30 somehow, thanks

let's say i had 90% on my exams, worth a total of 30%. Also, on homework i have a total of 85% which is worth a total of 40%. i want to know what i need to get on the final if the final is worth 30% of my grades? 90% is an A in the class. (.30)*(.90) + (.40)*(.85) + x = .90 i thought it...

so it converges, but no answer?

Determine whether the sequence a_n = \frac{1^2}{n^3} + \frac{2^2}{n^3} + ... + \frac{n^2}{n^3} converges or diverges. If it converges, find the limit. wouldnt it converge to 0?

Find the limit of the sequence whose terms are given by a_n = (n^2)(1-cos(5.2/n)) well as n->inf, cos goes to 1 right? so shouldn't the limit of this sequence be 0?

(y-7)^2 = 8(x-6) directrix x = _____ isnt the directrix just -p? 4p=8 p=2 so shouldn't the directrix be -2? but -2 doesn't work