Tuesday

September 2, 2014

September 2, 2014

Total # Posts: 124

**social studies**

Why do you think King Henry VIII insisted that everyone in England belong the Church of England?
*October 15, 2008*

**college physics**

i did not understand that! I NEED HELP!
*September 16, 2008*

**7th grade**

on wat
*September 16, 2008*

**math**

that is right i checked it!
*September 13, 2008*

**uop**

Benjamin O'Henry has owned and operated O'Henry's Data Services since its beginning ten years ago. From all appearances, the business has prospered. In the past few years, you have become friends with O'Henry and his wife. Recently, O'Henry mentioned that ...
*May 16, 2008*

**MATH**

I need to simply this equation, but I got stuck. h/(4-sqrt(16+h)) = y First, I multiplied (4+sqrt(16+h)/(4+sqrt(16+h) to both sides, and I ended up with h(4+sqrt(16+h)/-h. Is this correct? (I tried to graph both equations to see if I would get the same graph, but I didn't...
*August 8, 2006*

**MATH**

(x^2-2x+4)/ ((x-2)(x^2+4)) = y can this equation be simplified farther? I dont see an easy simplification.
*August 8, 2006*

**math**

I have to find an exact solutions on the interval [0,pi) for both of these problems-- 1) 2sin2x = sqrtx and 2) 1-2sqrt2 sinxcosx = 0 I know this much: 1) sin2x = (sqrt x)/2 = 2sinxcosx = (sqrt x)/2 What do I need to do next? 2) 1/(2 sqrt2) = sinxcosx = (sqrt2)/4 = sinxcosx = 2...
*August 1, 2006*

**factoring**

can this equation be factored further? y= x^4+2x^3+4x^2+8x+16 Not in the real number system. If you plot the function, you will see the minimum is at x=-1.1 (approx) and y is positive. At no x does the function equal zero, so there are no real roots, which means, no factors. ...
*July 26, 2006*

**math**

can this be simplified further? x^4+2x^3+4x^2+8x+16 Unless you are asking if the statement can be factored, the answer to your question is "no". Only variables with the same exponents can be added together. you mean this could be factored? how? dsfhsfhsrhsr
*July 25, 2006*

**math**

please help me simplify (x^3+8)/(x^4-16). thanks. The denominator is cleary the differerence of two squares (X^2-4)(x^2+4). That can again be factored to (x+2)(x-2)(x^2+4). The numerator can be factored using the trinomial simplification.. http://www.algebra-online.com/sum-...
*July 25, 2006*

**math**

Could you show me step by step how to find the inverse of this problem: log with base of 2 (x+1) Thanks! It is not clear what the problem is. You are given a function of x. Let's calli it y. Are you trying to come up with an equation for x in terms of y? If log(base2)of (x...
*July 21, 2006*

**math!**

about finding the inverse of y=(2x+1)/(x+3). should you multiply (x+3) on both sides and get y(x+3)=2x+1? then what would be your next step? This one is a little tricky because we we have a rational function. What we want to end up with is x(y) To do this you can interchange x...
*July 20, 2006*

**math**

how would you solve for y in this problem: x=(y^2+3y)^(1/3) I got: y= x^3/y - 3 ? and for this problem: x= 1/[(y^2+3y-5)^3] I got: y = y/x + 5/y - 3? However, I do not know how to solve for y explicitly. I don't think quadratic formula would work in these problems. Please ...
*July 20, 2006*

**sociology**

I happen to stumble accross this website and it is funny that it is still here after so many years...and that I am in Axia college and this is my week2 ethics course. It is very funny that the course has not changed in two years. I am also stumped on this question but have ...
*July 20, 2006*

**math**

how would you simplify y= (x^5-32)/(x-2)? I know that it would equal (x^5-2^5)/(x-2), but what's the next step? It wouldn't be just (x^4-2^4), would it? kristie, in general I think you can prove (x-a)|(x^n - a^n); read (x-a) divides the expression on the right of '...
*July 18, 2006*

**inverse**

the direction was to find f^(-1)x. f(x) = x^2+2x+1, x >= -1 as a final answer, I got: f^(-1)x = sqrt(x) - 1, x >= 0 is this right? I don't understand the restriction rules. It is right. If originally x>=-1, that puts f(x) >=0, and in the inverse function, this ...
*July 17, 2006*

**domain**

How do you find the domain of log(x^2-4)? and how about this one? e^(3x+4)? Please help me! Well kristie, you might recognize log(x^2-4) = log((x-2)(x+2)) and then use a property of logs to see it as a sum, i.e., log(ab)=log(a)+log(b). You are then expected to know that the ...
*July 17, 2006*

**interval**

Find the exact solutions on the interval [0,pi]. 2sin2x = sqrt(x) Please help me! kristie, this problem is somewhere between a calculus problem and an algebra one. I'm not sure what you're taking. If you see that the eq. is = to 4(sin2x)^2=x , then you might see where ...
*July 13, 2006*

**inverse**

my question was: find f^-1 (x). (this is asking me to find the inverse) f(x) = -(x-2)^2, x <= 2 how do I solve this problem?>> and you answered: If f(x)=-(x-2)^2 x<-2, then let y=f(x) - y = (x-2)^2 sqrt(-y)= x-2 x(y)= sqrt (-y) + 2 g(x)= sqrt(-x) + 2 and that is ...
*July 13, 2006*

**math**

how would you solve for y in this problem: ln(y-1)-ln2 = x +lnx ln(y-1)-ln2 = x +lnx Solve for y... ln[(y-1)/2]]=x + lnx take the antilog of each side (y-1)/2= e^(x+lnx) solve for y.
*July 11, 2006*

**inverse**

find f^-1 (x). (this is asking me to find the inverse) f(x) = -(x-2)^2, x <= 2 how do I solve this problem? find f^-1 (x). (this is asking me to find the inverse) f(x) = -(x-2)^2, x <= 2 how do I solve this problem?>> If f(x)=-(x-2)^2 x<-2, then let y=f(x) - y...
*July 11, 2006*

**math**

how would you simplify this equation: y = (x+3)/[(4-sqrt(16+h))] please help me! you have three variables. I am not certain "simplify" is an appropriate term here. ohhhh it was my mistake. I meant: y = h/[(4-sqrt(16+h))] y = h/[(4-sqrt(16+h))] rationalize the ...
*June 30, 2006*

**math**

how would you solve for y in this problem: x=(y^2+3y)^(1/3) would it equal: y= x^3/y - 3 ? and what about this problem: x= 1/[(y^2+3y-5)^3] would it equal: y = y/x + 5/y - 3? I would appreciate your help very much! <<how would you solve for y in this problem: x=(y^2+3y...
*June 27, 2006*