Math 95 Final Test Review help ASAP!?
I need help with these three math 95 problems for my final in less than 8 hours. These are all on my review so I have the answers but I cant figure out how to get them. If you dont know all of them, please answer the problem/problems you know. Thanks a bunch! Problems: 1: Determine the domain of the following function. k(x) = log(base 3)(x-7) 2: Perform the indicated operation, write results in simplest form. ( Just so you guys know, this is one of those cross canceling division problems, I couldn't write it the way its supposed to be due to the key board keys, but there is only supposed to be 1 division sign in the middle of all four problems. 4x^2-y^2 / 4x^2+2xy 7x^6y^5 / 14x^9y^3 Problem 3: The number of bacteria in a culture after t hours is given by N(t)= N(base 0)e^(0.028t) where N(base 0) is the initial number of bacteria in the culture. The initial number of bacteria is 1000. How long will it take for the number of bacteria to double? Give the answer correct to the nearest tenth. Answers 1:(7, positive infinity) 2: x^2(2x-y)/y^2 OR 2x^3-x^2y/y^2 3: 24.8 hours. THANKS!
Math problems!!! Please help me?
(1) The formula goes like this A = P ( 1 + r ) ^t P --> Principal or the initial amount r --> annual rate of increase (written as a decimal) t --> time in years A--> final amount or cost or whatever the problem is saying is increasing or decreasing y = 0.35 (2.3) * = 0.35 (1 + 1.3)* y --> the final amount 0.35 --> the principal ( initial amount) 2.3 = 1 + 1.3 , so " the r here" is 1.3 which when multiplied by a 100 = 130 % If "r " is positive then there is an increase in the amount of stuff every year If "r" is negative, then there is a decrease in the amount of stuff ever year log ^5 I am assuming means log of base 5 REMEMBER: THE WORD " LOG" MEANS " THE EXPONENT" and REMEMBER: 4 raised to -3 power = 1 / 4 ^3 x^-n = 1 / x^n log ^ 5 (1 / 25) in words means "TO WHAT EXPONENT MUST WE RAISE THE BASE (5) TO GET AN ANSWER OF 1/ 25 One way to do this: THINK OF HOW YOU CAN WRITE 1/ 25 as 5 to some power 1/25 = 1 / 5^2 = 5^ -2 log^5 (1/25) = log ^5 (5^-2) = -2 "To what power must you raise the base 5 to get 5 to the negative 2 power. " 5 must be raised to the -2 power to get 5 ^ -2 2b REMEMBER log (without the base shown means log of base 10) log (0.01) = log ( 1/100) = log (1 / 10^2) = log (10^ -2) = -2 "In words this means, " to what exponent must we raise the base 10, to get 10 to the negative second power" 3c. log^2 (1 / 128) = log^2 ( 1/ 2^7) = log^2 ( 2^-7) = - 7 "In words this means" " to what power must you raise the base 2, to get 2 to the negative 7th power. "
Math help!!!!please!?
16) 10 17) 13/60 18) 1 19) 9/20 20) D I'll do the rest in 20 minutes
Are there ways to solve math problems like this consistently?
One thing I want to warn you about. If the answer is a rational number, then a calculator will often only give you an approximate number. Like, in this case, the answer is 5/3 = 1.66666... Of course, a lot of calculators nowadays are smart enough to recognize that 1.666... is 5/3 and they will tell you the answer is 5/3 if you set them up correctly. If the instructor asks for an exact answer, then 1.666... is wrong. On a multiple choice test, you can use your calculator to see that 5/3 is equal to 1.666... so you can get around that. On an essay type question though, you will, and should, be marked wrong. The common base method (like in this case the common base is 2) is really not that hard to learn. Since common bases are almost always a prime number, like 2,3,5 or 7; or a small number, like 6, 8, or 10; it really doesn't take a whole lot of mental work to find the common base. I'm sorry to make things more complicated, but math can be like that some times.
Carbon 14. HELP ME FCKIN SOLVE THIS PLZ. FCKK FCK FCK FCK FCK.?
It's almost three half-lives. Thumbs up to kookoo for a pretty good estimation and thumbs down on everyone else. Edit: Okay, my curiosity got the better of me and I did the math. First thing to do is figure out how many half lives have gone by. This is a job for logarithms. 15% (or 0.15) is the proportion of original C14 remaining. Let x be the number of half lives. 2^x = (1/0.15) Okay, take the natural logarithm of both sides: x ln(2) = ln (1/0.15) divide both sides by ln(2) x = (ln (1/0.15)) / ln(2) Use your calculator x = 2.737 approximately So we have 2.737 half lives gone by to deplete the c-14 to 15% of its original. The half life of carbon 14 is 5730 years. 2.737 * 5730 years = 15683 years. So there's your answer: 15683 years.
Please help algebra 2 questions 10 points!!!!!! quick?
Logs are easy if you remember the equivalents http://www.ifomia.com/formula/math/logarithms.png For 1: Use basic algebra to solve for x. For 2: Re-write 1/81 in terms of 3. Hint 3 to what power equals 81? For 3: It's a rule you should remember... just google it then memorize it. It's better to learn and understand the rules, so you will know how to solve them on tests and such.