Wednesday

April 16, 2014

April 16, 2014

Number of results: 10,880

**another pre cal (logs)**

according to your basic laws of logs your expression is log3 ((3x-6)(x^2 - 4)/81)
*Monday, March 31, 2008 at 1:14am by Reiny*

**Pre Cal (logs)**

I need help with these logs 1: Log underscore4 16X 2: 2 Ln ( X(sqr. root)e 3:5^[2logUnderscore5(3x)] 4:log underscore2 ^[1280-logunderscore2 5) I don't understand how to do them thanks
*Sunday, March 30, 2008 at 11:25pm by Deb*

**another pre cal (logs)**

Thank you
*Monday, March 31, 2008 at 1:14am by Deb*

**more pre cal logs**

sorry I am so lost I missed the class that talked about logs since my sister was in the hospital giving birth to a still born baby.... now I am all lost... Expand the log Ln 3sqrootx^2y/x+3 (the 3 is part of the sqroot not infront like the 3rd power) I hope that makes since
*Monday, March 31, 2008 at 1:52am by Deb*

**more pre cal logs**

In your text or in your notes you should find 3 main rules for logs. 1. log(AB) = logA + logB 2. log (A/B) = logA - logB 3. log (A^n) = nlogA use them in this question. You did not use brackets to establish the correct order of operations here, so I will let you finish it. You...
*Monday, March 31, 2008 at 1:52am by Reiny*

**Pre-Cal**

solve for n 1 000 000 = 478 000(1.0628)^n you will have to use logs
*Sunday, June 12, 2011 at 9:03pm by Reiny*

**Pre-Cal**

Yes I am, but I don't understand it. I don't have a teacher to teach me. I do my work online, and I just need 2nd semester of pre-cal to graduate.
*Saturday, March 17, 2012 at 9:29pm by Adrianna*

**pre cal**

eh? plug in 6 for x. What do you get? If you're taking pre-cal, this should be a no-brainer.
*Wednesday, June 19, 2013 at 3:19pm by Steve*

**pre cal **

does anyone take pre-cal online ?
*Monday, April 16, 2012 at 6:44pm by Mecie *

**another pre cal (logs)**

write as a single log log(underscore3)(3x-6)-[log(underscore3)(x^2-4)+log(underscore3)81] thanks
*Monday, March 31, 2008 at 1:14am by Deb*

**pre-cal**

pre-cal? more like algebra I must be the review section for the x-intercept, set y=0: x=1 for the y-intercept, set x=0: y = -2
*Sunday, December 9, 2012 at 11:59am by Steve*

**pre-cal/trig**

if you get to cosx+sinx=1/square root of 2, well we are doing identities in my pre-cal class and cosx+sinx always equals 1, so it would be 1/square root of 2.
*Tuesday, March 4, 2008 at 7:44pm by emily*

**Pre Cal long problem( Can I see the work on this)**

You don't even need logs to do that question. From... 8(1+r) ^4=200 (1+r)^4 = 25 take the fourth root, (take the square root twice in a row) to get 1 + r = 2.236058 r = 1.236 the rate of return is 123.6% (wow) so $1000 invested at that rate for 3 years would have a value of ...
*Tuesday, March 18, 2008 at 4:16pm by Reiny*

**PRE-CALC still stuck**

Pre-cal? V=kT V=kr^2 I=V(1/R)
*Sunday, October 18, 2009 at 5:55pm by bobpursley*

**pre-cal... compound interest and modeling data**

1. To get the values of t you will need for your table, solve the equation (1+r)^t = 3.00 Hint: use logs. For r = 6% (1.06)^t = 3 t = log 3/log 1.06 = 18.85 years 2. I don't have a graphing calculator/utility. There might be some online application you can use. The product of ...
*Tuesday, December 11, 2007 at 10:55pm by drwls*

**Integrated Physics and Chemistry**

The specific heat capacity of copper is 0.09 cal/g°C. How much energy is needed to flow into a 10-gram sample to change its temperature from 20°C to 21°C? A)0.009 cal B)0.09 cal C)0.9 cal D)9 cal
*Thursday, April 21, 2011 at 10:29am by Ashley!!*

**math**

80 cal/g x 5 g 400g/cal or 400 cal/g not sure can any one help Neither. 80 cal/g x 5g = 400 cal. The unit g cancels to leave calories. x2+2x-15
*Monday, December 11, 2006 at 10:45pm by rena*

**chemistry**

How many calories are required to heat 25.0 g of platinum from 24.5 °C to 75°C (specific heat of platinum = 0.139 J/gK)? 1. 47 cal 2. 42 cal 3. 48 cal 4. 20 cal 5. 80 cal Thank you.
*Sunday, November 22, 2009 at 5:30pm by Danny*

**science**

the conversion factor for calories (cal) and joules (J) is: 1 cal = 4.184 J now to convert calories to joules, we multiply 4000 cal by 4.184 J / 1 cal so that the cal unit will be canceled, and the numerator will have the unit J: 4000 cal * 4.184 J / cal = 16736 J hope this ...
*Wednesday, October 26, 2011 at 3:09am by Jai*

**Pre-Cal**

Find the standard form of the equation of the ellipse. (Remember center is midpoint between either foci or vertices) (1) 9x^2+4y^2+36x-24y+36=0 (2) Vertices: (0,5)(0,-5) Passes through the point (4,2) Centered at the origin
*Sunday, April 19, 2009 at 6:33pm by Cal*

**Math**

Logs are stacked up in apile as shown in the figure. The top row has 15 logs and the bottom row has 21 logs. How many logs are in the stack? a1= 1+14=15 an= 7+14=21 s1=(n/2)(a1+an) (n/2)(15+21) = 18 How do I figure out the n?
*Tuesday, May 4, 2010 at 4:37pm by Abbey(Please help)*

**Pre-Calculus**

I'm trying to finish up my pre-cal homework and I am stuck on 2 problems... ***determine the intervals over which the function is increasing, decreasing, or constant*** 32. f(x)=x^2-4x 33. f(x)=√x^2-1 (that's square root of x^2-1) thank you so much for the help =)
*Saturday, September 17, 2011 at 2:40pm by Cristina*

**Pre-Calculus**

I'm trying to finish up my pre-cal homework and I am stuck on 2 problems... ***determine the intervals over which the function is increasing, decreasing, or constant*** 32. f(x)=x^2-4x 33. f(x)=√x^2-1 (that's square root of x^2-1) thank you so much for the help =)
*Saturday, September 17, 2011 at 3:48pm by Cristina*

**Science**

A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000 cal. D) 5250 cal. Can someone please help me solve...
*Monday, September 26, 2011 at 1:29am by Anonymous*

**physics **

A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.) Answer A) 1250 cal. B) 4000 cal. C) 5000 cal. D) 5250 cal. Can someone please help me solve...
*Monday, September 26, 2011 at 5:17pm by Allison*

**Chemistry**

You don't need the heat of vaporization to do that calculation. It takes 10g*1Cal/C g*10C = 100 Cal heat release to reduce the temperature ot 0C, and another 80 Cal/g*10 g = 800 Cal to freeze it. The sum is 900 Cal
*Saturday, April 17, 2010 at 12:35am by drwls*

**Nautical/Maths**

By the way, in the old days like when I went sailboat racing, we had to use logs of trig functions to do these products by adding and subtracting logs because we did not have calculators. This was a total mess and resulted in complicated things like using half the angles so ...
*Friday, January 4, 2008 at 3:49am by Damon*

**math - natural logs**

Find the EXACT solution to e^(1-x) = 4^x You must use logs, not just a calculator.
*Monday, October 31, 2011 at 10:10pm by BJ*

**Pre. Cal.**

yes
*Thursday, May 14, 2009 at 10:43pm by Reiny*

**Pre cal**

OK. Thanks. =]
*Saturday, August 29, 2009 at 6:12pm by Anonymous*

**Pre-Cal**

Thank You
*Saturday, March 27, 2010 at 5:38pm by Abbey*

**Pre-Cal**

Thank You!!
*Sunday, March 28, 2010 at 12:54pm by Hannah*

**Pre-Cal**

Thank You
*Sunday, April 4, 2010 at 5:30pm by Abbey*

**Pre Cal**

b)
*Monday, July 25, 2011 at 3:04pm by Anonymous*

**Pre Cal**

why and how do you know
*Monday, July 25, 2011 at 3:04pm by Ashley*

**Pre Cal**

Thank you!!!
*Monday, February 13, 2012 at 11:06pm by Anonymous*

**Pre-Cal**

thanks
*Saturday, March 17, 2012 at 9:29pm by Adrianna*

**Pre-cal**

thank you
*Wednesday, April 18, 2012 at 10:10am by sandra*

**Pre-cal**

to do what?
*Thursday, December 13, 2012 at 5:57pm by Steve*

**college maths**

Mathematically, you can set up the problem as follow: 232 Cal/1.5 Cup = x Cal/1 cup 1x = 232/1.5 x = 154.7 Cal Another way, this is a no brainer question given the choices below. With 1.5 cup, you know it has 232 Cal, so with only 1 cup, you know the answer should be less than...
*Wednesday, December 15, 2010 at 2:16pm by Terry*

**Math**

you will have to use logs 4( 7^(x-2) ) = 8 divide both sides by 4 7^(x-2) = 2 take logs of both sides log (7^(x-2) ) = log 2 by rules of logs (x-2)(log 7) = log 2 x-2 = log 2/log 7 x = 2 + log2/log7 = appr 2.3562
*Thursday, January 23, 2014 at 10:06am by Reiny*

**Chemistry**

1.24E7 J x (1 cal/4.184J) = 2.98E6 cal. 12 oz x 2.98E6 cal/80,000 cal = ?oz Gatorade.
*Saturday, April 20, 2013 at 7:55pm by DrBob222*

**Pre-Cal**

What are bearings??
*Tuesday, February 5, 2008 at 12:34pm by Daisy M.*

**Pre Cal**

thanks so much!
*Tuesday, February 19, 2008 at 1:01pm by Kindell*

**Pre-Cal**

xy=2 {{x^2=3+y^2}}
*Tuesday, October 21, 2008 at 6:16pm by Ash*

**Pre Cal.**

correct.
*Tuesday, March 31, 2009 at 9:28am by bobpursley*

**pre cal--check**

yes
*Tuesday, March 31, 2009 at 8:27pm by Reiny*

**Pre Cal.**

è is theta.
*Tuesday, April 21, 2009 at 3:12pm by Jamie*

**Pre Cal.**

correct!
*Tuesday, April 21, 2009 at 3:12pm by Reiny*

**Pre-Cal**

(-2,1) and (2,1) are the vertices.
*Saturday, May 2, 2009 at 7:37pm by Hal*

**Pre Cal**

Solve for x 2^x=8^x-3
*Tuesday, December 22, 2009 at 2:53am by Raina*

**Pre-Cal**

Everywhere BUT 0
*Thursday, March 25, 2010 at 10:22pm by Damon*

**Pre-Cal :)**

You're very welcome!
*Saturday, March 27, 2010 at 5:38pm by MathMate*

**Pre-cal**

I agree
*Saturday, May 8, 2010 at 12:23am by Reiny*

**Pre Cal**

What is the inverse of g(x)= 1/2(x-1)^3-4
*Monday, October 25, 2010 at 1:17am by Katie*

**Pre Cal**

40
*Saturday, January 9, 2010 at 5:50pm by Anonymous*

**Pre Cal**

this is wrong
*Sunday, July 24, 2011 at 8:20pm by Dalidah*

**pre-cal**

no "t" in formula
*Monday, January 16, 2012 at 8:07pm by Steve*

**Translation**

It is pre-cal
*Friday, March 23, 2012 at 4:25pm by Sebastian*

**Pre-cal**

123694587
*Thursday, May 6, 2010 at 7:21pm by Anonymous*

**Pre-Cal**

#3 = B 95
*Saturday, March 17, 2012 at 10:21pm by Ryan*

**pre cal**

what is the question?
*Wednesday, February 13, 2013 at 5:08pm by bobpursley*

**Pre cal**

second
*Thursday, January 30, 2014 at 8:42pm by shelly *

**Pre cal**

how do you find A?
*Thursday, January 30, 2014 at 8:42pm by shelly *

**pre cal**

mmmhhh? (0+3)^2 = 3^2 = 9
*Thursday, January 30, 2014 at 8:53pm by Reiny*

**pre cal**

zero
*Thursday, March 6, 2014 at 9:43am by Damon*

**pre cal**

s400
*Sunday, September 4, 2011 at 11:14pm by Halib*

**Chemistry**

Do you know the conversion factor? 1 cal = 4.184 joules and I get 4565 cal. Is that a capital C for cal. In nutrition, especially in the old days, there was a big calorie (actually a kilocalorie or Calorie) and a small calorie (a calorie). If we round the 5 to the even number...
*Friday, October 30, 2009 at 7:22pm by DrBob222*

**Precalculus**

.5=e^(-20t) I am assuming this involves logs but we did't really learn logs. Anyone care to walk me thru with a calculator. Have the Ti-83
*Thursday, March 13, 2008 at 7:08pm by Trevor*

**Pre-Cal (again)**

If: f(x)= 3/x^2-1 g(x)= x+1 What is the domain of f, g, and f o g?
*Saturday, September 20, 2008 at 9:19pm by Crystal*

**Trig/Pre-Cal**

1 - - x > 0 x
*Sunday, September 7, 2008 at 11:18pm by Brooke*

**pre cal**

What would cos2theta=-1/2 be???
*Saturday, November 15, 2008 at 2:45pm by becca*

**pre cal**

what is the sum of e or 1n
*Sunday, January 11, 2009 at 6:07pm by ash*

**Pre-Cal**

Find f X g 3 sqrt of x+1 X x^3+1
*Thursday, February 26, 2009 at 11:32pm by Dan*

**Pre-Cal**

if 2^x=3 whats does 4^-x equal
*Thursday, February 26, 2009 at 11:32pm by Dominique*

**Pre-Cal.**

Graph it on your calculator.
*Thursday, March 26, 2009 at 9:16am by bobpursley*

**Pre Cal.**

all 3 are correct.
*Monday, June 15, 2009 at 6:29pm by Reiny*

**Pre Cal.**

All correct!
*Thursday, August 13, 2009 at 12:45am by MathMate*

**Pre Cal.**

Both correct!
*Thursday, August 13, 2009 at 12:49am by MathMate*

**Pre Cal.**

none of them are correct.
*Saturday, August 15, 2009 at 4:31pm by bobpursley*

**Pre-Cal**

Simplify e^3ln2-1
*Saturday, January 9, 2010 at 11:02pm by jamie*

**Pre-Cal**

1/sec = cos
*Friday, March 5, 2010 at 5:42pm by bobpursley*

**Pre-Cal**

yes, to your second part
*Wednesday, March 10, 2010 at 11:12pm by Reiny*

**Pre-Cal(Please help)**

Error
*Tuesday, March 23, 2010 at 1:05pm by Mark*

**Pre-Cal(urgent)**

Nevermind.
*Wednesday, March 31, 2010 at 11:06pm by Hannah*

**Pre-cal**

both are right.
*Sunday, May 9, 2010 at 7:59pm by bobpursley*

**pre-cal**

(576^1/2+512^1/3)^1/3
*Monday, May 31, 2010 at 7:53pm by Soto*

**pre-cal**

16
*Thursday, September 2, 2010 at 6:06pm by Ms. Sue*

**L Pre-Cal**

yes that was a typo, my bad!
*Wednesday, January 26, 2011 at 7:43pm by Norah*

**Pre Cal**

please explain your b and a.
*Monday, July 25, 2011 at 3:04pm by Ashley*

**Pre cal**

Thank you sooooo much !!!
*Sunday, October 23, 2011 at 8:28pm by Basim*

**Pre Cal**

cos^2+tan^2=1
*Sunday, March 25, 2012 at 7:56pm by Trevion*

**pre-cal**

thank you very much
*Wednesday, April 18, 2012 at 10:52am by sandra*

**Pre-Cal**

YOU SHOULD KNOW HOW TO DO THIS!
*Wednesday, March 25, 2009 at 2:29pm by YOU'RE BAD AT MATH*

**pre cal**

7tan^2(pheta)-2=3
*Thursday, June 28, 2012 at 10:48pm by Maria*

**pre cal**

(y^2)^-4y^8 simplify
*Thursday, August 30, 2012 at 11:20pm by jennifer*

**pre-cal**

That's what I thought but was not for sure. thanks!!
*Saturday, October 13, 2012 at 5:21pm by ladybug*

**pre cal**

by definition, tan(t) = b/a
*Sunday, February 17, 2013 at 2:18pm by Reiny*

**Pre cal**

ix^2-2x+i=0
*Sunday, March 17, 2013 at 6:21pm by Anonymous*

**Pre-Cal**

tan-1=43\9.3
*Wednesday, May 1, 2013 at 8:01pm by Anonymous*

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