All Answers 358080 Answers

1. -6 + (14 + 2[60 - 9(1 + 3)]) = 56

2. (6² - 2²) / 8 = 4

3. 7/3 + 3 (3/4 - 2/3)² = 2.35

4. 6/7 + 6(1/2 - 1/3)² = 1.02

Hope this helps! 

-RedGlasses

Pretty good. Thanks for asking! :D

Pretty good, you?

What is 45+23+45

45+45= 90

90+23= 113

45+23+45= 113

TheXSquaredFactor... I Forgot all about that... I am so sorry for the misunderstanding... I take back what I said... I'm sorry 

I'm assuming this is :

50  = 2x^0.8        divide both sides by 2

25  = x^0.8   

25  = x^(8/10)

25 = x^(4/5)      raise both sides to the 5/4 power

25 ^ (5/4)   = x   ≈  55.9

Well, remember when I said that it has no common factors? Well, I was wrong. 

It turns out that the GCF of  7428564 and 999999 is not 1. 

It turns out that the GCF of both of those numbers is 142857. Wow!

Therefore, \(\frac{7428564}{999999}\div\frac{142857}{142857}=\frac{52}{7}=7\frac{3}{7}\)

That looks so much nicer. I think you'd agree!

 -2m - 5m - 8 = 3 + (-7) + m   combine like terms

-7m - 8  =  -4 + m        add 7m, 4 to both sides

-4  = 8m       divide both sides by 8

-4/8  = -1/2  = m

5 + 4 = 9

(-5) + 4 = -1

(-5) + (-4) = -9

5 + (-4) = 1

Two regular pentagons and a regular decagon, all with the same side length, can completely surround a point, as shown.

An equilateral triangle, a regular octagon, and a regular -gon, all with the same side length, also completely surround a point. Find .

regular 24-gon

(cotx/sinx-tan^2x- tanx/cos+2x/cos^2x)dx

NOTE I have added an x after cos as I think that you forgot it.  :)

\(\int (\frac{cotx}{sinx}-tan^2x-\frac{ tanx}{cosx}+\frac{2x}{cos^2x})dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\~\\ -----\\ \int \frac{cotx}{sinx}dx=\int \frac{cosx}{sin^2x}dx=\int cosx(sinx)^{-2}dx=-(sinx)^{-1}+c_1=-cosecx+c_1\\ \int tan^2x dx=\int \frac{sin^2x}{cos^2x} dx=\int\;(sec^2x-1)dx=tanx-x+c_2\\ \int\frac{ tanx}{cosx} dx=\int\frac{ sinx}{cos^2x} dx=(cosx)^{-1}+c_3=secx+c_3\\~\\ \text{So far I have }\\ =\:-cosecx-(tanx-x)-(secx)+\int\frac{2x}{cos^2x}dx\\ =\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx \)

\(=\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx\\ consider\;\;\;\int\frac{2x}{cos^2x}dx\\ =\int(2x(sec^2x)dx\\ \text{Solve using integration by parts}\\ u=2x \qquad v'=sec^2x\\ u'=2 \qquad v=tanx\\ \int\;uv'dx=uv-\int(vu')dx\\ =2xtanx-\int(2tanx)dx\\ =2xtanx-2\int(\frac{sinx}{cosx})dx\\ =2xtanx+2ln(cosx))+c_4\\~\\ \text{So my final answer is}\\ =\:-cosecx-tanx+x-secx+2xtanx+2ln(cosx)+c\\ \)

Here's a method to convert any repeating decimal to a fraction. 

The first step is to set it equal to a variable; I'll use the standard one, x.

\(7.428571428571...=x\)

Next step is to get only one portion of the repeating portion to the right hand side of the equation. I will demonstrate this:

\(\begin{align} 742857.428571&=100000x\\ 7.42857142857...&=x\\ \end{align}\)

Subtract the equations together to get the following:

\(7428564=999999x\)

Divide by the coefficient of x on both sides.

\(x=\frac{7428564}{999999}\)

This is your final answer. I do not believe the numerator and denominator have any common factors. 

It's probably 9. Probably. It might be 1. But let's just go with 9. 

Please include your question in the body of your post .. not just in the title :/

x-y=5      (1a)          becomes       x=5+y  (1b)

-4x+4y=20  (2a)     becomes         -x+y=5     becomes      x=y-5 (2b)

Put these together and we get

5+y=y-5

That is never going to happen :(

So there are no solutions.   Yes it is very sad :((

What so funny ?? 

AAWWWWW my goodness!!...hahaha,

gosh!!!...man!!!....i'm sitting here crying with laughter...Melody!!..thank you very very much!..you are great!!

58 is 58.0 and you can add as many more 0s as you like.

When you subrtract or add then you have to line up the points :)

So the 8 has to go above the 3  

The 8 and the 3 are both in the units column :)

Hi Juriemagic,

It is good to see you again :)

what you have done is correct but you would neer leave it like that :)

S=P(1+rt) 

\(S=P(1+rt) \\ t=(\frac{S}{P}-1)\div r \\ t=(\frac{S}{P}-1)\times \frac{1}{r} \\ t= \frac{1}{r}(\frac{S}{P}-1)\\ t= \frac{S}{Pr}-\frac{1}{r}\\ \)

I dont even know what to say ginger. I was not trying to be arrogant, nor i was trying to attack her or any other moderator. I thought the way she presented her answer was a bit complicated and hard to find. That's it. I already mentioned that there are easier ways to answer this, and i thought that maybe presenting JUST the table with all the combinations is complicated for that. Im sure melody can take a little bit of criticism.

You also corrected me about "this is the best way i know"-

1. I was talking about the general case where every die has a different number of sides.

2. This is the best way i know (knew).

3. When i said "simple" i meant a single binomial coefficient that wi solve the question like the simple binomial coefficient that is the solution when there's no upper bound.

(But be my guest, find a general, simple formula for the general case, it's worth trying isnt it ;))

________________

It seems as if you're mostly trying to defend melody, whenever someone says something about her that you dont like. I didnt say melody is a bad moderator, nor that she's stupid for not using the way i did. I pointed out the fact that there is a simpler solution that i THOUGHT she should've presented. 

But when im complimenting jz1234 for coming up with an idea that was closer than everyone else in the thread? (Again, this might seem like im attacking melody. You can calm down, ginger! Take a deep breath, drink a glass of water, because im not attacking her, im mentioning a fact)- no, dont even THINK about complimenting him, because he didng give the EXACT formula for the solution.

Now, you may not believe me, but i knew you'll criticize me for my post. I knew you'll call me a blarney banker, i knew you'll defend melody. This leaves two options:

1.Im the oracle!

2. You're just getting predictable.

tl;dr: grow up

Interestingly, a staircase diagram shows this function is cyclic once it reaches x = -1:

.

f(x)  is defined only for  0 ≤ x ≤ 1  , and  f(x)  =  ax + b  for constants  a  and  b  where  a < 0  .

Since  a  is negative, the smallest possible value for  f(x)  will be when  x = 1  .

f(1)  =  a(1) + b  =  a + b

Then, the largest possible value for  f(x)  will be when  x = 0  .

f(0)  =  a(0) + b  =  b

The smallest possible value for  f(x)  is  a + b  , and the largest possible value for  f(x)  is  b  .

So......the range for  f(x)  is  [ a + b ,  b ]  .

I don't really know what it means to make the division complete, but if  a = -1  , then...

\(\frac{x^2+ax-2}{x-2}\,=\,\frac{x^2-x-2}{x-2}\,=\,\frac{(x+1)(x-2)}{x-2}\,=\,x+1\qquad\text{, when}\,x\neq2\)

Nice, Gh0sty  !!!