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Are we speaking plain decimals here, or money problems?

How do you type all those symbols Chris? The mejority of those I don't even know how to locate, and that doubles on not knowing what some of them mean, hopefully this web can help me with that!

This  has a slant asymptote  of y = x  and a vertical asymptote of x = 2

See the graph, here : https://www.desmos.com/calculator/klzevxjfpe

log₂(x)=2

Easier than you may tnink, D_r  !!!

In exponential form, we have that  the log base raised to the "2"  on the right side of the equation  =  x     .....so......

2^2  =  x

4  = x

Prety easy, huh???

Any one ? 

f(x)=(-2x)/(x^2-x-12)

This will be undefined whenever  ( x^2 - x - 12)  = 0

So.....factoring, we  have

(x - 4) ( x + 3)   = 0

And the function is  undefined  when x = 4 or  x = -3

So.....4 is the larger number not in the domain

`

Example :

cos ( 60°)  =  1/2    so

[ cos (60° ] ^5  =   (1/2)^5   =  1/32

(a + b + c)^3   = 4^3

a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3 =  64

[a^3 + b^3 + c^3] + 3 [ a^2b + a^2c +b^2a + b^2c + c^2a + c^2b] + 6abc  = 64

[a^3 + b^3 + c^3] + 3a^2(b + c) + 3b^2(a + c) + 3c^2(a + b) + 6abc  = 64

[a^3 + b^3 + c^3] + 3a^2(4 - a) + 3b^2(4 - b) + 3c^2(4 - c) + 6abc  = 64

-2[ a^3 + b^3 + c^3] + 12[a^2 + b ^2 + c^2]  + 6abc  =  64     ...... divide by  -2

[ a^3 + b^3 + c^3] - 6[ a^2 + b^2 + c^2] - 3abc  = - 32      (1)

And  (a + b + c)^2  = 16

a^2 + 2 a b + 2 a c + b^2 + 2 b c + c^2  = 16

[a^2 + b^2 + c^2] + 2[ ab + bc + ca]  = 16

[a^2 + b^2 + c^2]  + 2 [ 7 ]  = 16

[a^2 + b^2 + c^2]  + 14  = 16

[a^2 + b^2 + c^2]  = 2     sub  this into  (1)

[ a^3 + b^3 + c^3] - 6[ 2 ] - 3abc  = - 32

[ a^3 + b^3 + c^3] - 12 - 3abc  = - 32

a^3 + b^3 + c^3 - 3abc  =  - 20

g(7) = |1-7| =|-6| = 6

f(g(7)) = f(6) = 6^2 + 4(6) = 36+24=60 

Since you cannot have a ln of zero or negative numbers:

2x-20> 0

2x>20

All 'x' such that :  x> 10    is the domain

6x−y+3z = 28  (1)

5x−2y+2z = 28  (2)

−3x−y+4z = -17   (3)

Multiply  (1) by -2......add to  (2)  and we have

-7x  - 4z  = -  28   (3)

Multiply  (3) by -2......add to  (2)  and we have

11x - 6z  = 62  (4) 

Multiply  (3) by -6  and  (4) by 4

42x + 24z  = 168

44x - 24z  =  248        add these

86x  = 416       divide both sides by 86

x  = 416/86   =  208/43

Using (4)  to find z, we have

11(208/43) - 6z  = 62

2288/43 - 62*43/43  = 6z

-378 /(6 *43)  = z

-63/43  = z

And using (1)  to find y

6x−y+3z = 28

y = 6x + 3z -28

y = 6(208/43) + 3(-63/43) -[28 *43] / 43

y = -145/43

 make x the subject of the formula y=x+a

y=x+a subtract a from both sides

x = y - a

Divide equation 3 by 43:
{6 x - y + 3 z = 28 | (equation 1)
0 x - 7 y - 3 z = 28 | (equation 2)
0 x+0 y+z = (-63)/43 | (equation 3)
Add 3 × (equation 3) to equation 2:
{6 x - y + 3 z = 28 | (equation 1)
0 x - 7 y+0 z = 1015/43 | (equation 2)
0 x+0 y+z = -63/43 | (equation 3)
Divide equation 2 by -7:
{6 x - y + 3 z = 28 | (equation 1)
0 x+y+0 z = (-145)/43 | (equation 2)
0 x+0 y+z = -63/43 | (equation 3)
Add equation 2 to equation 1:
{6 x + 0 y+3 z = 1059/43 | (equation 1)
0 x+y+0 z = -145/43 | (equation 2)
0 x+0 y+z = -63/43 | (equation 3)
Subtract 3 × (equation 3) from equation 1:
{6 x+0 y+0 z = 1248/43 | (equation 1)
0 x+y+0 z = -145/43 | (equation 2)
0 x+0 y+z = -63/43 | (equation 3)
Divide equation 1 by 6:
{x+0 y+0 z = 208/43 | (equation 1)
0 x+y+0 z = -145/43 | (equation 2)
0 x+0 y+z = -63/43 | (equation 3)
Collect results:
Answer: |x = 208/43        y = -145/43           z = -63/43

.....36  45 54 63 72 81 .....etc

They are both driving the same final speed, so the second car will never catch up with the one that had a head start.

95/(7+12) x 7 = 35  units  (one side)

95/(7+12) x 12 = 60 units (other side)

What is the reciprocal of 4 1/2 ?

Anything times 0 is 0...Therefore 

3x=0 

x=0 

Plug in the answer to see if it is correct 

3(0)=0 

x=0

a x 1/2 =b..........(1)

a + 1/2 =b..........(2)

From 2 above we have:

a = b - 1/2, we sub into (1)

(b -1/2) x 1/2 =b

1/2b - 1/4 = b

- 1/4 = b - 1/2b

- 1/4 = 1/2b divide both sides by 1/2

b = - 1/4 x 2

b = - 1/2 sub into (2) above and we have:

a + 1/2 = - 1/2

a = -1/2 - 1/2

a = - 1

Awesome if you knew the answer why did you ask cphill lol

A number besides 0. That both multiplied by 0.5 and added to 0.5 gives the same result

a x 1/2 =b

a + 1/2 =b

Solve the following system:
{a/2 = b | (equation 1)
a + 1/2 = b | (equation 2)
Express the system in standard form:
{a/2 - b = 0 | (equation 1)
a - b = -1/2 | (equation 2)
Swap equation 1 with equation 2:
{a - b = (-1)/2 | (equation 1)
a/2 - b = 0 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{a - b = -1/2 | (equation 1)
0 a - b/2 = 1/4 | (equation 2)
Multiply equation 1 by 2:
{2 a - 2 b = -1 | (equation 1)
0 a - b/2 = 1/4 | (equation 2)
Multiply equation 2 by 4:
{2 a - 2 b = -1 | (equation 1)
0 a - 2 b = 1 | (equation 2)
Divide equation 2 by -2:
{2 a - 2 b = -1 | (equation 1)
0 a+b = (-1)/2 | (equation 2)
Add 2 × (equation 2) to equation 1:
{2 a+0 b = -2 | (equation 1)
0 a+b = -1/2 | (equation 2)
Divide equation 1 by 2:
{a+0 b = -1 | (equation 1)
0 a+b = -1/2 | (equation 2)
Collect results:
Answer: |a = -1            and             b= -1/2

There are infinite solutions to your problem, such as:

3, 43, 23 so that: 3+43 =2x23, or

5, 53, 29 so that: 5+53 =2 x 29 or

19, 43, 31 so that: 19+43 =2 x 31.......etc.

a^3 + b^3 + c^3 -3abc, when ( a+b+c) = 4, & (ab+bc+ca)=7

Is there a minus in front of the a^3 or not???

Try expanding (a+b+c)^3 and see what you get.  I have not done it but I expect it all will become obvious :)