Tuesday

December 6, 2016
Total # Posts: 20

**Geometry**

36*4= 144 144/9=16 X= 16 peices

*August 31, 2011*

**Algebra**

How do you find the zeroes of the 1/2x^3-x? I factoried it so it would be 1/2x(x^2)-x. The one zero would then be zero, which matches with my book's answer. But I have no idea how my book got +/- the sq. root of two for the other one. Can you demonstrate, please?

*November 3, 2009*

**Algebra**

I had to find the zeroes of (2t-3)/(t+5). I thought one of the answers could be -5, in addition to 3/2. But my book says -5 isn't an answer. Why?

*November 3, 2009*

**Algebra**

How do you find the zeroes of x/(9x^2-4)? I thought the answer would be the sq. root of 4/9, but my book says the answer is zero. Why is this?

*November 3, 2009*

**Algebra**

I am trying to figure out the zeroes of 2x^2-7x-30. I did this graphically in a way that matched up with the answers in my book, however, I couldn't get it by distributing. I tried (2x+-10)(x+3) but that didn't get -7x in the middle as logic told me it should. How ...

*November 3, 2009*

**Algebra**

I have to find the inverse function of f(x)=3-4x. I got y=3-x/4 for the function. I then have to plug it back into the function and verify that it is the inverse. So I wrote (3-4(3-x))/4. This resulted in (-3+x)/4=x. I tried isolating it to get x=x but I just can't get it...

*November 3, 2009*

**Algebra**

(f/g)(-1)-g(3) when f(x)=x^2+1 and g(x)=x-4. I got -1^2/1-4-(3-4), which became 2/-3 + 1. I multiplied the 1 by -3, which resulted in -3/-3 or 3/3. 2/-3+3/3 is 1/3. But my book says the answer is 3/5. Where did I go wrong?

*November 3, 2009*

**Algebra**

f(x)=2x-5 g(x)=2-x. I had to find f/g(x). I wrote 2x-5/2-x but this was marked incorrect. How would you do it then?

*November 2, 2009*

**Algebra**

How can you tell an equation's real or imaginary roots just from its graph, without knowing the equation? I have to look ath the graph of a parabola, absolute value, and a negative third degree equation and do so? Are there any links that could explain this as I can't ...

*November 2, 2009*

**Algebra**

f(2)=5 f(-2)=1, I have to write the linear function for this. I got 5-1/2- -2=4/4=1 then I plugged it in so it would be y-5=1(x+2) which became y=x+7, then f(x)=x+7. The answer key, however, says f(x)=x+3. What did I do wrong?

*November 2, 2009*

**Algebra**

I'm not sure how to find the domain for the 4th root of (1-x^2). Can you break down the process for me please?

*November 1, 2009*

**Healthy Decision Making**

Could you define the context of "risks" and "injury"? Is this for a unit on drugs, athletic injury, or just decisions in general?

*November 1, 2009*

**sci**

Think of what appliances, electronics, and utilities require energy. You can do this just by looking around your house and seeing what has a plug. Then, think of the consequences of what would happen if these things are always running. Who would get affected? Would it cause ...

*November 1, 2009*

**Civics**

First, look in the dictionary and make sure you understand the definition of lobbying, or use the glossary in your book. Then, think of who might lobby and why they might do so. Write the result on here and I'll give you my opinion.

*November 1, 2009*

**Algebra**

The function 5x^2+2x-1's domain is all real numbers, according to the answer key. However, I don't know how to factor this to prove it. Could you show me? The function the sq. root of y-10 has a domain of all numbers so that y is greater than or equal to zero. But I ...

*November 1, 2009*

**algebra**

In the magic square shown (I'll type it out), the sum of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by v,w,x,y, and z. Find y + z. Here's what the magic square looked like. I'll type semi-colons to seperate ...

*February 25, 2009*

**Question**

I finished it up like I said with elimination and got y = -3. Is this correct? If not, I'll try it the way you suggested with elimination the whole way through. Thanks, Liese

*February 25, 2009*

**Algebra**

Find all solutions to the following system of equations: x + 3y + 2z = 6 -3x + y + 5z = 29 -2x - 3y + z = 14 This just did not work out for me. Here is my work: I first solved Equation #1 for x. x = -3y - 2z + 6 Then, I did substitution, and replaced x for -3y - 2z + 6 in ...

*February 25, 2009*

**Algebra**

Solve the following system of linear equations: 2x - 3y + 6z = -12 5x + 2y - 8z = 29 7x + 6y + 4z = 49 First, I did elimination for the first two equations. I multiplied the first equation by two, and the second equation by 3. 4x - 6y + 12z = -24 + 15x + 6y - 24z = 87 That ...

*February 25, 2009*

**math**

Uggh... I am doing the same thing right now, and not fully understanding either, but I'll try to explain. What I was told is to solve the first equation for x or another variable. In this case, it would be easier to do z because it doesn't have a number next to it. ...

*February 25, 2009*

- Pages:
**1**