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x = number

-x = opposite of number

5(-x) + 56 = 2x

-5x +56 = 2x

56 = 7x

8 = x

m=(y+x)/2 for y    cross multiply

2m =y +x             subtract x from both sides

2m - x = y,           or,

y =2m - x

100/ big number. Then take that answer and multiply it by the smaller number you gave us.

the answer is 4

Take log of both sides to get

(3-x) LOG 2 = (2x+3) LOG (1/3)      LOG2 =.3010    LOG(1/3) = -.4771

.90308 - .301x = -.9542x - 1.4313     Now just solve for 'x'

2.33438= -.6532x

x= -3.5738

(x/3) + (y/2) = 14

(x/2) + (y/5) = 10

Solve the following system:
{x/3 + y/2 = 14 | (equation 1)
x/2 + y/5 = 10 | (equation 2)
Swap equation 1 with equation 2:
{x/2 + y/5 = 10 | (equation 1)
x/3 + y/2 = 14 | (equation 2)
Subtract 2/3 × (equation 1) from equation 2:
{x/2 + y/5 = 10 | (equation 1)
0 x+(11 y)/30 = 22/3 | (equation 2)
Multiply equation 1 by 10:
{5 x + 2 y = 100 | (equation 1)
0 x+(11 y)/30 = 22/3 | (equation 2)
Multiply equation 2 by 30/11:
{5 x + 2 y = 100 | (equation 1)
0 x+y = 20 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{5 x+0 y = 60 | (equation 1)
0 x+y = 20 | (equation 2)
Divide equation 1 by 5:
{x+0 y = 12 | (equation 1)
0 x+y = 20 | (equation 2)
Collect results:
Answer: |x = 12                   and                    y = 20

You can't really ''solve'' this problem because we don't know the actual sum of the equation, are you sure you don't mean ''simplify'' and not ''solve''?

Solve for x over the real numbers:
2^(3 - x) = 3^(-3 - 2 x)

Take reciporicals of both sides:
2^(x - 3) = 3^(2 x + 3)

Take the natural logarithm of both sides and use the identity log(a^b) = b log(a):
log(2) (x - 3) = log(3) (2 x + 3)

Expand out terms of the left hand side:
log(2) x - 3 log(2) = log(3) (2 x + 3)

Expand out terms of the right hand side:
log(2) x - 3 log(2) = 2 log(3) x + 3 log(3)

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

Divide both sides by log(2) - 2 log(3):
Answer: |x = (3 log(2) + 3 log(3))/(log(2) - 2 log(3))= -3.573804392782666.........

There is because it's part of an assigment I received in math class.

Also I really meant  "Find S" as in "find the ensemble of solutions for this problem".

If ''how much water is in each lake?'' refers only to Lake Superior, Michigan and Huron, this should be the answer.

The question gives us some information on the lakes' sizes which we can use to find the answer.

S-1720=M 

S-(1720+330)=S-2050=H

Now we do some basic algebra and find the value of S.

S+(S-1720)+(S-2050)=4930

3S-3770=4930

3S=8700

S=2900

Now we use S' value to find the value of M and H.

2900-1720=1180=M

2900-2050=850=H

Answer: Lake Superior contains 2900 cubic miles of water, Lake Michigan contains 1180 cubic miles of water, and Lake Huron contains 850 cubic miles of water.

By "Find S" you meant "Find x".

Tony, it looks like you dreamt this one up yourself!! Why? Because there is NO SOLUTION for x!!!!!>

Let's say

Lake Mich = x

Lake Superior  is then 1720 + x

Lake Huron is  x-330

Theay all ad up to 4930

x + (1720+x) + (x-330) = 4930

x = 1180 = Mic

Super = 1720 + 1180 = 2900

Huron = 1180-330 =850

melody gettin fed up

correction

(25y^6)^1/4=

Do not know what format you are looking for, but here are some possibilities:

= (25^1 * y^6)^1/4

= 25^(1*1/4)  *  y^(6*1/4)

=25^(1/4) * y^(6/4)

=(5^2)^1/4 * y^3/2

=5^(2/4) * y^3/2)

=5^(1/2) * y^(3/2)

=sqrt (5) *  y^(3/2) 

=\(\sqrt{5}\)\(\sqrt[2]{y^3}\)

21/3 = 7 yards

In case you have not yet learned about derivatives (to find the slope):

We can identify the minimum or maximum value of a parabola by identifying the y-coordinate of the vertex. You can use a graph to identify the vertex or you can find the minimum or maximum value algebraically by using the formula x = -b / 2a. This formula will give you the x-coordinate of the vertex.

the parabola equation we had was:

Area = 1/2  px - x^2    

        = -x^2 + 1/2  px         a = -1    b = 1/2  p

the MAXIMUM of the parabolic area equation will be     @  x = -b/2a =  (-1/2p) / 2(-1)  = 1/4 p     and again we have the value of a side which gives maximum area is   x = 1/4 p     so the rectangle is a square....

I would add that  OPPOSITE   and ADJACENT   are NOT the HYPOTENUSE .....

You put in ADJACENT/OPPOSITE     This is COTANGENT

OPPOSITE/ADJACENT  = TANGENT   ( T-O-A)

= 10/19.72  = Tangent  

Try THAT and let me know what you find !

Let

x= one side length

2x= length of TWO sides

P-2x = length of OTHER TWO side

(P-2x)/2 = length of ONE other side

Area = side 1 x side 2

        = x * (P-2x)/2

        = 1/2Px -2x^2/2

        = 1/2 Px -x^2       

This is an upside down 'U' shaped parabola.....we are asked to find the MAXIMUM value of this area equation

   This will be where the slope = 0

   First derivative defines slope          SLOPE= 1/2  P - 2x    Set this equal to ZERO

  1/2 P - 2x = 0  

2x=1/2 P

x = 1/4 P         So the maximum area is acheived when x = 1/4 P   

the SECOND side will be ( P- 2x)/2  = (P- 2(1/4P))/2 = 1/2P / 2 = 1/4P    The same as the FIRST side

  If all of the sides are equal (which we just shown) the rectangle with the biggest area and perimeter P  is a SQUARE !

8^(1/3) = \(\sqrt[3]{8}\) = 2

If you mean 3 fourths:

3/4=0,75

0,75+0,05=0,8

If you mean three fours:

4*3+0,05=12+0,05=12,05