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by theshadow
Mon Sep 12, 2016 4:48 pm
Forum: Beginning Algebra
Topic: Solving One Step Equations: (-3/5)a = 27
Replies: 2
Views: 128

Re: Solving One Step Equations: (-3/5)a = 27

The first problem that we need help with is: \dfrac{-3}{5}\, a\, =\, 27 In the second step, they are multiplying both sides by 5 which we understand, but how are they getting -3a as the answer on the left after multiplying by 5? \begin{align} \dfrac{-3}{5}\, a\, &=\, 27 \\ (5)\, \dfrac{...
by theshadow
Mon Sep 05, 2016 3:07 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Solve using induction n! < ((n+1)/2)^n
Replies: 1
Views: 157

Re: Solve using induction n! < ((n+1)/2)^n

n! < ((n+1)/2)^n if n>2 Your subject line says that you need to "solve using induction", but induction is a form of proof, not of solution. You can learn about what induction is, and how to apply it, in this lesson . Where are you stuck in the process? Please show ALL of your work so far....
by theshadow
Wed Jul 06, 2016 5:08 pm
Forum: Beginning Algebra
Topic: Order of operations isn't working.
Replies: 3
Views: 405

Re: Order of operations isn't working.

Q = 1,500 – 4P + 5A + 10I + 3PX P = $400 A = $20,000 I = $15,000 PX = $500 pemdas Q = 1500 – 4 * 400 + 5 * 20000 + 10 * 15000 + 3 * 500 Q = 1500 – (4*400) + (5*20000) + (10*15000) + (3*500) Q = 1500 – 1600 + 100000 + 150000 + 1500 Q = 1500 – 253100... Without any grouping symbols telling you to, wh...
by theshadow
Sat Jun 11, 2016 1:37 pm
Forum: Geometry
Topic: Find the values of the opposite sides
Replies: 1
Views: 480

Re: Find the values of the opposite sides

Hello Everyone, I have to solve this quadrilateral problem and no idea where or how to start top left corner is A bottom left corner is B bottom right corner is C top right corner is D Find a way to get the values of CD and DA AB - BC = DA - CD AB - CD = DA - BC AB + DA = BC + CD I'd start by namin...
by theshadow
Fri May 27, 2016 10:47 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: Solve for X (Expressions with Roots)
Replies: 3
Views: 610

Re: Solve for X (Expressions with Roots)

Solve for x: ux^a = x The answer they give is: Step 1: ux^a = x Step 2: x^(a-1) = (1/u) Step 3: x = (1/u)^((1/(a-1)) I understand until that part. The next step, I don't understand. Step 4: x = u^(1/(1-a)) How do you get from Step 3 to Step 4? They used exponent rules and flipping a subtraction. . ...
by theshadow
Thu May 12, 2016 6:00 pm
Forum: Calculus
Topic: How to prove f(x) = (cx-c*x+x-1)/(2x-x-1), where 1<c<2 is decreasing function?
Replies: 3
Views: 312

Re: How to prove f(x) = (cx-c*x+x-1)/(2x-x-1), where 1<c<2 is decreasing function?

How to prove the following function is decreasing function? f(x)\, =\, \dfrac{c^x\, -\, cx\, +\, x\, -\, 1}{2^x\, -\, x\, -\, 1} \mbox{with }\, 1\, <\, c\, <\, 2\, \mbox{ and }\, 0.5\, <\, x\, <\, 1 This one probably works just like the other one, so use the same method. What did they use f...
by theshadow
Thu May 12, 2016 5:46 pm
Forum: Calculus
Topic: How to prove f(x) = [(x-1)*cx-2*x*cx-1+x+1]/(2x-x-1) is increasing function?
Replies: 1
Views: 231

Re: How to prove f(x) = [(x-1)*cx-2*x*cx-1+x+1]/(2x-x-1) is increasing function?

Can you help me improve the following function f(x) is an increasing function? Sure, but we need to see what you've done first. How far did you get, and where are you stuck? . f(x)\, =\, \dfrac{(x\, -\, 1)\, c^x\, -\, 2xc^{x-1}\, +\, x\, +\, 1}{2^x\, -\, x\, -\, 1} \mbox{ with }\, 1...
by theshadow
Thu May 12, 2016 5:39 pm
Forum: Intermediate Algebra
Topic: Doing a comprehensive review, need fraction help...
Replies: 1
Views: 267

Re: Doing a comprehensive review, need fraction help...

The problem is: 2/a 2 - 3/ab + 4/b 2 This is just an expression. Until they give you instructions, there is no problem do resolve. (answer: 2b 2 - 3ab + 4a 2 / a 2 b 2 ) My guess is that they forgot to tell you to "convert to a common denominator and combine". With other fractions that ha...
by theshadow
Fri Apr 01, 2016 3:09 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: solving for x, given exponentials: 4e^2x + 5e^3x = 15
Replies: 7
Views: 870

Re: solving for x, given exponentials: 4e^2x + 5e^3x = 15

denicarl wrote:if I have 4e^2x + 5e^3x = 15
how do I solve for x?

What you wrote means this:

a.

Is that right? Or did you mean this?

b.

denicarl wrote:I tried ln 4 +2x+ln5+3x=ln 15 but that doesn't work.

How did you get to this step?

Thanks.
by theshadow
Fri Apr 01, 2016 3:08 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: solving for x given: 4e^-.2x + 3e^-.3x = 5
Replies: 4
Views: 446

Re: solving for x given: 4e^-.2x + 3e^-.3x = 5

denicarl wrote:how do I solve for x given: 4e^-.2x + 3e^-.3x = 5

The way you've written it means this:

a.

Did you mean this?

b.

Thanks.

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