All Answers 358468 Answers

thank you so much you don't know how much it took me to find out about this you just made me take a big step in my college :)

Since  log(G)  =  365.8532 / 200     --->   log(G)  =  1.829266

To find G: use the antilog (which means raise to the power of 10):  G  =  10^1.829266

--->  G  =  67.5  (approx)

Eureka!

4!!+(4+4)/4

Where "!!" stands for double factorial.

$$1\frac{1}{4}*4\frac{1}{2}=$$

$$(1+\frac{1}{4})*(4+\frac{1}{2})=$$

Left, outer, inner, right:

$$4+\frac{1}{2}+1+\frac{1}{8}$$

Multiply(*) both the numerator (top) and the denominator (bottom) in fraction nr. 1 by 4 to get a common denominator:

$$4+\frac{4}{8}+1+\frac{1}{8}$$

$$5+\frac{5}{8}$$

Remember that: $$a\frac{b}{c}= \frac{ac +b}{c}$$, we get:

$$5\frac{5}{8}=\frac{40+5}{8}=\frac{45}{8}=5.625$$

We should call ourselves the "Dazed and Confused" forum      

4/4+4/4+4/4+4/4+4/4+4/4+4/4+4/4+4/4+4/4=10

In maths it is called factorial

6 factorial = 6! = 1*2*3*4*5*6 

$${\mathtt{6}}{!} = {\mathtt{720}}$$

It is heavily used in probabability   

$$\\-3=2x^2-5x\\
0=2x^2-5x+3\\
$multiply to 6, add to -5 numbers are -2 and -3$\\
$change -5x into -2x-3x$\\
0=2x^2-2x-3x+3\\
0=2x(x-1)-3(x-1)\\
0=(2x-3)(x-1)\\
2x-3=0 \qquad or \qquad x-1=0\\
x=3/2\qquad \quad or \qquad x=1\\$$

How about: 

4 + √4 +√4 +√4

Technically speaking you've used two's since √x = ²√x

The expression is ambiguous, but I'll assume you mean:

$$\frac{3/4(-8)}{2-3}$$  

The order of operations are: 

1. Solve what's in the paranthesis

2. Multiply (*) and divide (:)

3. Add (+) ans subtract (-)

We get:

$$\frac{3/4(-8)}{2-3} = \frac{3/-32}{-1}$$

Multiply both the numerator (top) and denominator (bottom) with -1

$$\frac{3/32}{1}=3/32 = 0.09375$$

2+2 = 5-1

hgjghjgjgjjkhkjkhj

log^a3 + log^a8 - log^a2 as a single logarithms

I assume this is supposed to be

loga^3 + loga^8 - loga^2    if so, we have...

log [ (a^3 * a^8) / a^2) ] =

log ( a^11 / a^2) =

log a^9 =

9log(a)

------------------------------------------------------------------------------------------------------------------------

 solve 2^(x+1) = 3^x to 2 d.p.

Take the log of both sides

log 2^(x + 1) = log 3^x     and by a property of logs, we have

(x + 1) log 2 = x log3       simplify

x log 2 + log 2  = x log3     rearrange

x log 2 - x log 3  = - log 2     factor

x (log 2 - log 3) = - log 2      divide both sides by (log 2 - log 3)

x = (-log 2) /(log2 - log3) = about 1.71

IV in Roman notation ;)

2^-1 = 1/2 = 0,5

Did that help ?

U r welcome :)

okay thank you

d = 13       C = ?

Remember the formula:

$$C = pi*d$$

So we get: 

C ≈ 3,14* 13 = 40,82

C ≈ 40,82

2x +3y = 34 

2x + 3(5x) = 34  (from y = 5x)

2x + 15x = 34

17x = 34 

x = 2

y = 5x

y= 10

Welcome to the club.......!!!!.....most of us on here are "Dazed and Confused"

0.0016931 = 1.6931* 10^-3.

So we get:

(1.6931* 10^-3)(10⁶) = 1.6931* 10^6-3 = 1.6931* 10³ = 1693,1

Scientific notation is on the form: a*10^b.

Dispose of all the trailing zeroes, and set the period after one single digit: We get: a = 4.81

The exponent is one less than the number of digits. Therefore, "b" is 13.

So we get: 4,81 * 10¹³

2log(x - 3)  =  1 + log(x) + log(2) - log(5)

log(x - 3)² - log(x) - log(2) + log(5)  =  1    (multiplier goes inside as an exponent)

log[ 5(x - 3)² / (2x) ]  =  1                         (subtraction --->   division; addition   --->  multiplication)

5(x - 3)² / (2x) = 10                                  (raise both sides as exponents upon 10)

(x - 3)² / (2x)  =  2                                   (divide by 5)

(x - 3)²  = 4x                                           (multiply by 2x)

x² - 6x + 9  =  4x                                     (multiply out)

x² - 10x + 9  =  0

(x - 9)(x - 1)  =  0

x  =  9   <---- answer                 x = 1   <--- won't check

Do you mean?:

$$(6*7)(\frac{1}{7})$$

We see that one of the factors is equal to the denominator (bottom) in the fraction. They cancel out, and we get:

(6*7)(1/7) = (6*1)(1/1) = 6 

As many - or as few - as you would like......!!!! There are infinite possibilities....