All Answers 358504 Answers

20(.4) + x    =  (20+x)(.75)

Solve for x = 28  cars must be added.

Probably to check and confirm your work.

Guests are notorious for posting wrong answers, errant logic, and just plain blarney and bullshit. 

There a few longtime guests who perpetually do this very thing. I call them the BBs. I can usually recognize them because they all write with quills. One of them often commands his computer to write with a quill –it never seems to run out of ink. 

GA

Your script code is incorrect.  Probably because my sample code above is incorrect. Despite checking it several times after refreshes, one of the sub-compilers eventually assimilated part of it.  Communicating the correct compiler code is a “Catch 22” That is, when I post the code script the compiler encodes (or partly encodes) it and it disappears from view. Modifying it so it presents normally and does not compile in whole or part is tricky.

All of the this presentation should remain visible indefinitely.

<span class="math-tex">\({}\)</span>

Despite being modified so it presents and does not compile, this can be copied and pasted and it will work as it is supposed to (unless it buggers up again).   You can always use the LaTeX box with {}, which is what this command does anyway.

GA

DOG

DGO

GOD

GDO

ODG

OGD        (if you have to use all of the letters)

Multiply all of the dimensions to find the volume:

(x+5)(x-5) = x^2 - 25

(x^2-25)(x^2 +25)= x^4 - 625

this volume has to be less than 700 units^3

x^4 -625 < 700

x^4 < 1325

x< 6.033          so integers   1,2,3,4,5,6   would work for x       BUT you cannot have a negative (or zero) dimension so ONLY the integer 6 will work....so only ONE integer value will work for volume <700 units^3

Six possibilities.

Where is the rest of your question?  

Thanks for the help, but I'm still having trouble solving for the answer. I don't really understand the video. 

You could also just use the rational root theorem

\(\text{Count up the additional sundaes using less ingredients}\\ N_3 = 8 (3\cdot 2 + 4 \cdot 3 + 4 \cdot 2) = 208\\ N_2 = 8(4 + 3 + 2) = 72\\ N_1 = 8\\ N_1+N_2+N_3 = 288\)

I have been stuck on this for a long time. 

i =Internet stock

1.1 i = internet stock after year

10 000 = copper stock

9/10 (10000) = copper stock after year

1.1 i + 9/10 (10000)  = 1.06(i +10000)

..04 i = 1600

i = 40000

At the end of the year 1.1i = 1.1 (40000) =$ 44000

The answer is \(-\frac{22}{5}\) but if you want the full explantion you get get that here 

https://www.algebra.com/algebra/homework/Graphs/Graphs.faq.question.986526.html  

I don't know if this is the correct way to post a answer because I have not done it. 

Hope it Helps!!

l1  is re-written as   y = -5/8 x -9/8     slope m is  -5/8

l2 perpindicular slope is     -1/(-5/8) = 8/5

y = 8/5 x + b     sub in point   10,10

10 = 8/5 (10) + b

b = 10 - 80/5 = -30/5

-30/5 +8/5=  -22/5

\(((x-a)(x-b)(x-c) =x^3 +  x^2 (-a-b-c)+x (a b+a c+b c)-a b c \\ 52=a+b+c\\ 581 = ab + ac + bc \\ k = abc\)

\(\text{Solving this is a pain but conceptually straightforward.}\\ \text{Start by knowing one of them must be 2, as the sum of 3 odds is odd and 52 is even}\\ \text{The roots are $2, ~13,~37$}\\ k = 2 \cdot 13 \cdot 37 = 962\)

 span> 

\(\text{This is a test $x^2 + y^2 = r^2$} \)

wooo!

Except the editor chews on the enable line and leaves span> there.

THX I never figured out how to use the desmose graphing calculator. but THX for the help. 

Thanks for the help! I got the answer to the problem after a lot of work. 

Meloody, why did you post this answer which is exactly like the answer I posted yesterday?  I'm not being accusatory or anything like that; I genuinely want to know why.  Thanks. 

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Here I have built the graph for you.

You can work it out from the picture.

\(\sqrt{(x-3)^2+(y+4)^2} + \sqrt{(x+5)^2+(y-8)^2}=20\\~\\ \)

Sorry, i have no idea. 

Here you can see the simplified version

By observation, we can see that x = 1 is a root because (1)^3 - 2(1) + 1 = 1 - 2 + 1 = 0

If x = 1, then (x - 1) = 0

Assuming x^3 - 2x + 1 = 0 can be written as (x+a)(x+b)(x+c) = 0, where -a, -b, and -c are the roots, then we can divide the equation by the root we already know to simplify it.

(x^3 - 2x + 1) / (x - 1) = x^2 + x - 1

Use the quadratic equation to find the roots of x^2 + x - 1

x = (-b +/- (b^2 - 4ac)^(0.5)) / (2a) = (-(1) +/- ((1)^2 - 4(1)(-1))^(0.5)) / (2(1)) = -0.5 +/- (5^0.5) / 2

The other 2 roots are x = -0.5 + (5^0.5) / 2 ~= 0.6 and x = -0.5 - (5^0.5) / 2 ~= -1.6

The largest root is x = 1, so let r = 1, s = 0.6, t = -1.6

r^3 - s - t = (1)^3 - (0.6) - (-1.6) = 2

2 is the maximal value!

Solve for x:
x^3 - 2 x + 1 = 0

The left hand side factors into a product with two terms:
(x - 1) (x^2 + x - 1) = 0

Split into two equations:
x - 1 = 0 or x^2 + x - 1 = 0

Add 1 to both sides:
x = 1 or x^2 + x - 1 = 0

Add 1 to both sides:
x = 1 or x^2 + x = 1

Add 1/4 to both sides:
x = 1 or x^2 + x + 1/4 = 5/4
Write the left hand side as a square:
x = 1 or (x + 1/2)^2 = 5/4

Take the square root of both sides:
x = 1 or x + 1/2 = sqrt(5)/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:
x = 1 or x = sqrt(5)/2 - 1/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:

x = 1 or x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2
1^3  - [sqrt(5)/2 - 1/2] - [ -1/2 - sqrt(5)/2] = 2

It does not make sense to me either :/

I have not  worked this out

BUT

you can think of A,B and C being on an xyz space such that 

A(12,0,0)    B(0,0,12)     and    C(0,6,0)

and T os the origin (0,0,0)

Then I'd try doing what this video teaches.