Let x be the number of motorcycles and y be the number of cars...each car has 4 wheels and each motorcycle has 2...so we have

4x + 2y = 310    we have more variables than equations, so there will be multiple integer solutions

These square root equations can be a little "sticky"..... here we go......

√(2x + 4) - √(x + 1)   =    √( 3x - 19)      first, square both sides....this gives

2x + 4 - 2√[(2x + 4)(x + 1)] + x + 1  = 3x - 19     simplify 

3x + 5 - 2√[(2x + 4)(x + 1)] = 3x - 19    leave the root on the left and move everything else to the right side by subtrating 3x, 5 from both sides....this gives us

- 2√[(2x + 4)(x + 1)] = -24      divide both sides by -2

√[(2x + 4)(x + 1)]  = 12           square both sides

(2x + 4)(x + 1) = 144               simplify

2x^2 + 6x + 4 = 144               subtract 144 from both sides

2x^2 + 6x - 140   = 0              divide through by 2

x^2 + 3x - 70   = 0                  factor

(x + 10) (x - 7) = 0               setting each factor to 0, we have posible solutions of x = -10 and x = 7

Notice that -10 makes all three original roots negative and we can't take a square root of a negative number (this is known as an "extraneous" solution)

Check x = 7 in the original problem....

√(2(7) + 4) - √(7 + 1)   =    √( 3(7) - 19)

√(18) - √(8)   =    √( 2)

3√(2) -  2√(2)   =    √( 2)

√( 2) = √( 2)

Yep...that works right out !!!

Be sure to ask if you have questions....these problems are a little difficult.....

6x^3+38x^2+18x+2=0

ignore the last post. I thought but means different.

I think your example doesnt work all the time. If I just always leave the negative out side, I will always get a negative answer for all powers. It cannot be right example.

when does it remain inside of the brackets

When is there a 1 put outside brackets?

for n+(-1)squared i just6 take out the minus first?

how do I know if w = 0. It is in a function, maybe at some points it equals 0.

For -3 squared, two negative is positive isn't it, even amount of powers is positive, uneven is odd

you have erros in your example

1/2pi squareroot(pi*0.0004^2/1000*1.5*(2*10^-13)

enter like this if you want what is below.  

1/2*pi*sqrt((pi*0.0004^2)/(1000*1.5*2*10^-13))

$$\frac{ \pi }{2} \sqrt{\frac{\pi*0.0004^2}{1000*1.5*2*10^{-13}}}$$

You can do this on the site calc on  the home page.  

These are good questions - welcome to web2.0calc forum :)

Any real number divided by itself = 1       except for 0       0/0  is undefined

So w/w = 1 so long as w is not equal to 0   

NOW

$$\\(-3)^2=-3*-3=9\\
but\\
-3^2=-1*3^2=-1=9=-9\\\\
$I know this is confusing but think about this question.$\\
20-3^2=20-9=11\\
$See how the - sign did not get squared ?$\\
$If I can help more or explain more let me know. $$$    

You've probably already figured this one out - simply multiply both sides by 31.7

2^(-3) is the same as 1/(2^3), in other words, just flip negative exponents between the numerator and denominator.

2 divided by 3^4 

is the same as

2 times 3^(-4).

(2^2014)-(2^2013)=(2^2013)*2-(2^2013)*1

So factoring, (2^2013)*(2 - 1)=2^2013 * 1.

Verbally, you are subracting half the original value from that value.  Which will leave half the original value remaining.

f(x)  =  [2 - 3x] / [5 - 2x]

f(a)  =  [2 - 3a] / [5 - 2a]  =  3

Multiply both sides by 5 - 2a:

             [2 - 3a]  =  3[5 - 2a] 

                2 - 3a  =  15 - 6a

Add 6a to both sides:

               2 + 3a  =  15

                     3a  =  13

                       a  =  13/3

You can use the distance formula:  d  =  √[ (x2 - x1)² + (y2 - y1)² ]

(x1, y1) = (0, 5)     --->     x1 = 0   and   y1 = 5

(x2, y2) = (3, 10.5)   --->   x2 = 3   and   y2 = 10.5

d  =  √[ (3 - 0)² + (10.5 - 5)² ]   =   √[ (3)² + (5.5)² ]   =  √[ 9 + 30.25 ]    =   √[ 39.25 ] 

The square of a number is that number times itself.

6²  6 x 6  =  36

If the problems is stated as:  ax² + bx + c  =  0,

the quadratic formula is:        x  =  [ -b ± √(b² - 4ac) ] / (2a)

When graphed, this is a hyperbola. The question is:  What y-value is impossible?

If you find the lim(x→∞) [3x - 7] / [x + 1], you will be able to answer the question.

Rewrite  [3x - 7] / [x + 1]  =  [3 - 7/x] / [1 + 1/x]   (by dividing all terms by x0

--->    lim(x→∞) [3 - 7/x] / [1 + 1/x]  =  3.

To show that 3 is impossible, try to solve the equation:  3  =  (3x - 7) / (x + 1)

     Multiply both sides by  x + 1     --->     3(x + 1)  =  3x - 7     --->     3x + 3  =  3x - 7

     --->  3  =  -7   Impossible!

Thus, the range is all numbers but 3.

For numbers smaller than -1, g(x) will have a negative number under the radical sign, will be undefinded for real numbers.

However, f(x), for values less than -1, both the numerator and the denominator will be negative, giving a positive answer, and will then be defined.

The functions will be equaivalent if x > -1; for x ≤ -1, f(x) will be defined but g(x) won't be.

E(r, r, 3)  =  625  =  r x r^3  =  r^4

                     625  =  r^4

Find the fourth root of both sides:

                    625^(1/4)  =  r

                           --->   r  =  5

You are wasting your time. You wouldn't know what to do with it even if you had one.  

All numbers can be substituted for x except 0, therefore:  -∞ < x < 0  or  0 < x < ∞

--->   {-∞, 0}  ∪  {0, ∞}     --->   (-∞, 0)  ∪  (0, ∞) 

f(x)  =  3x² - 4x

f(k - x)  =  3(k - x)² - 4(k - x)  = 3(k² -2kx + kx²) - 4k + 4x  =  3k² - 6kx + 3x² - 4k + 4x 

for f(x)  =  f(k - x)     ---     3x² - 4x  =   3k² - 6kx + 3k² - 4k + 4x 

The 3x²  on the left side matches the 3x² on the right side, so we can ignore them leaving:

 - 4x  =   3k² - 6kx - 4k + 4x 

On the right side: 3k² - 4k is a constant; there is no equaivalent on the left side, so 3k² - 4k  =  0

--->     k(3k - 4)  =  0       --->     So either k = 0  or 3k - 4  =  0   --->  k = 0  or  k  =  4/3

What we have left:  -4x  =  -6kx + 4x

                                -4x  =  (-6k + 4)x

The only way that this can be true is if  -4  =  -6k + 4, or  k = 4/3

Answer:  k  =  4/3

3/4 should be greater than 5/7