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Wed Jun 04, 2014 3:06 pm
Forum: Intermediate Algebra
Topic: Simplifying equation of a parabola
Replies: 1
Views: 1759

### Simplifying equation of a parabola

Hi, The text defines the equation of a parabola as: \sqrt{x^2+(y-p)^2} = y+p It goes on to say: By squaring and simplifying we get x^2 = 4py . I'm trying to recreate the steps they took to get from the first form to the second. I start by removing the radical sign by multiplying both sides by \sqrt{...
Sat Mar 29, 2014 2:13 pm
Forum: Geometry
Topic: Question about similar triangles
Replies: 3
Views: 8315

### Re: Question about similar triangles

I posted this over on StackExchange and got the following answer: For two similar triangles, the ratio of the height of the first triangle to the height of the second triangle is equal to the ratio of the base of the first triangle to the base of the second triangle. http://math.stackexchange.com/qu...
Fri Mar 28, 2014 7:01 pm
Forum: Geometry
Topic: Question about similar triangles
Replies: 3
Views: 8315

### Question about similar triangles

My text states the following: A_1 = \frac{1}{2}*h_1*b A_2=\frac{1}{2}*h_2*e \frac{A_1}{A_2}=\frac{\frac{1}{2}*h_1*b}{\frac{1}{2}*h_2*e}=\left(\frac{h_1}{h_2}\right) \left(\frac{b}{e}\right)=\left(\frac{b}{e}\right) \left(\frac{b}{e}\right)=\left(\frac{b^2}{e^2}\right) I'm not following how they're g...
Sun Dec 04, 2011 8:07 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Finding an Exponent Variable
Replies: 2
Views: 2423

### Re: Finding an Exponent Variable

Do you see where to go now? Ahh...so: \frac{2^6}{3^3} + \frac{2^6}{3^2} = \frac{2^a}{3^b} \frac{2^6}{3^3} + (\frac{2^6}{3^2}*\frac{3}{3}) = \frac{2^a}{3^b} \frac{2^6}{3^3} + \frac{3*2^6}{3^3} = \frac{2^a}{3^b} \frac{64}{3^3} + \frac{192}{3^3} = \frac{2^a}{3^b} \frac{256}{3^3} = \frac{2^a}{3^b} \fra...
Sun Dec 04, 2011 6:49 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Finding an Exponent Variable
Replies: 2
Views: 2423

### Finding an Exponent Variable

Hi again, The problem states: If (\frac{3}{4})^{-3} + (\frac{8}{3})^2 = \frac{2^a}{3^b} find a and b The answer is a=8 and b=3, but I'm stumped as to how they got there. It seems like I need to turn the right-hand side into an addition expression. I can convert it to a product by expressing 3^b as 3...
Sun Dec 04, 2011 6:22 am
Forum: Advanced Algebra ("pre-calculus")
Topic: Negative Exponent Problem
Replies: 2
Views: 1817

### Re: Negative Exponent Problem

Instead, you'd need to convert the "2" to "32":
I love you!

$\frac{1}{x}*\frac{x^4}{x^4} + \frac{1}{x^5}=\frac{x^4+1}{x^5}$

Thank you,
mb
Sun Dec 04, 2011 12:33 am
Forum: Advanced Algebra ("pre-calculus")
Topic: Negative Exponent Problem
Replies: 2
Views: 1817

### Negative Exponent Problem

Hi, The problem states: "Write the following expression with only positive exponents and in simplest form. Variables are not equal to zero". x^{-1} + x^{-5} The answer is: \frac{x^4+1}{x^5} From the initial expression I get: \frac{1}{x} + \frac{1}{x^5} It's here that I get stuck. The next step seems...
Sun Nov 27, 2011 12:25 am
Forum: Advanced Algebra ("pre-calculus")
Topic: Understanding Exponents
Replies: 4
Views: 2928

### Re: Understanding Exponents

Ok, so that would be:
$25^{b+1} = 5^{b+2}$, right?

Still stumped...
Sat Nov 26, 2011 6:05 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Understanding Exponents
Replies: 4
Views: 2928

### Understanding Exponents

If $25^{b+1}= x$ and
$5^b = y$

then express x in terms of y.

The answer it gives is:

$x=25y^2$

I'm struggling to understand the formula or operation(s) one would go through to answer this.

Help?
mb
Sun Nov 13, 2011 11:43 pm
Forum: Beginning Algebra
Topic: Help Simplifying an Equation
Replies: 2
Views: 2384

### Re: Help Simplifying an Equation

Ahhh...

$2a_1+nd-1d = 2a_1+(n-1)d$

Thank you!

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