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Let's look at this again......and we will remove the last 3 constraints.....these three just tell us that  the solution will occur in the first quadrant  where x  ≥ 2  and   y ≤ 11

Look at the graph now : https://www.desmos.com/calculator/upxf7i7tbd

If I read the graph correctly.....[a real leap of faith]...there appear to be only two corner points that satisfy the constraints....they are

(8,2)   and (9,4).......but only one will maximize  4x + y....and that occurs at (9,4)

JK it equals no swag at all everybody except me and some of my friends have no swag at all thats the answer NO SWAG AT ALL

3x+3x+2x=180°

           8x=180°   | :8

            x=22,5°

3x=67,5°

2x=45°

I think its a C

x³ + x -10 < 0  ........  this is probably best done graphically

https://www.desmos.com/calculator/kcwl3qnk6g

It appears that the solution  is  ( -inf, 2)

log base 3 (x) = log base 5 (125)

2(LOG₂X)+3=0    subtract 3 from each side

2(LOG₂X)   = -3     divide both sides by 2

(LOG₂X)  = -3/2     In exponential terms, we have that

2^-3/2  =  x         means the same as

(1/2) ^3/2   = x    ....so....

x = [(1/2)^3] ^(1/2)   =    √(1/8)    =   (1/2)(1/√2)  =   1 / [2√2]

The max will occur at a "corner point" but this is very hard to identify with a graph because there are so many "overlapping"  graphs....

WolframAlpha gives the following solution : 

https://www.wolframalpha.com/input/?i=maximize+4x+%2By+given+3x%2By++%E2%89%A4+31+,++2x%2By++%E2%89%A4+23,+++2x-y++%E2%89%A4+14,+++x%2B6y++%E2%89%A5+20,+++y+%E2%89%A4+11,+++y+%E2%89%A5+0,+++x+%E2%89%A5+2

It appears that the max occurs at (9, 4)   and the max is  4(9)  + 4   = 40

Here's the Desmos graph : https://www.desmos.com/calculator/opl1cuchfu

Maybe you can find the max corner point much easier than I can.......!!!!!

I don't know how to delete a thread but please don't waste time here. I made a typo in the initial equation and that is why sh*t hit the fan....

I think you might want to take your dog to the vet...

1+1=3

Come on no one what's to f**k

15*X = 124

X = 124/15

X = 8.27

42/[1+2*(12-9)]

= 42/[1+2*(3)]

= 42/[1+6]

= 42/7

= 6

speed = distance/time

so speed = (3/5 miles)/(3/8 hours)

speed = (3/5)*(8/3) --> 3's cancel out

speed = 8/5 miles per hour

Huge thanks to everyone!

I mostly try to do it myself first but then validate with what you guys provide.

35 pounds/114 pounds per day = 0.3 days

-8*8 = -64 + 81*-10 = -810

-74

Don't know what you've graphed either Melody.

Anyway,

\(\displaystyle e^{x}-3e^{-x} + 2=(e^{2x}+2e^{x}-3)/e^{x}\)

\(\displaystyle =(e^{x}-1)(e^{x}+3)/e^{x}\),

and that will be greater than zero whenever

\(\displaystyle e^{x}-1 > 0\) .

That is, whenever x > 0.

what is 40/7

\(\frac{40}{7}=5\frac{5}{7}=5.71428\overline{571428} \ periodic\)

             !

sorry,I typed wrong. 

(x/8)-(x/2)=6+(6x/16)

\(\frac{x}{8}-\frac{x}{2}=6+\frac{6x}{16} \)

\(\frac{2x-8x}{16} =\frac{16* 6+6\color{red}x}{16} \)  

\(-6x-6x=96\)

\(-12x=96\)

\(x=-8\)     !

Solve for p over the real numbers:
abs(6 - 7 p) = 13

Split the equation into two possible cases:
6 - 7 p = 13 or 6 - 7 p = -13

Subtract 6 from both sides:
-7 p = 7 or 6 - 7 p = -13

Divide both sides by -7:
p = -1 or 6 - 7 p = -13

Subtract 6 from both sides:
p = -1 or -7 p = -19

Divide both sides by -7:
Answer: |p = -1       or       p = 19/7

1+2=795

Solve for X:
3 + (2 log(X))/(log(2)) = 0

Bring 3 + (2 log(X))/(log(2)) together using the common denominator log(2):
(3 log(2) + 2 log(X))/(log(2)) = 0

Multiply both sides by log(2):
3 log(2) + 2 log(X) = 0

Subtract 3 log(2) from both sides:
2 log(X) = -3 log(2)

Divide both sides by 2:
log(X) = -(3 log(2))/2

-(3 log(2))/2 = log(1/2^(3/2)) = log(1/(2 sqrt(2))):
log(X) = log(1/(2 sqrt(2)))

Cancel logarithms by taking exp of both sides:
Answer: |X = 1/(2 sqrt(2))

Keep it clean here