geno3141

First, remove the parentheses by using the distributive property:

     2(y - 7)  =  2·y - 2·7   =   2y - 14

     6(y + 1)  =  6·y + 6·1  =  6y + 6

7y - 2(y - 7)  =  6(y + 1) - 4

7y - [2y - 14]  =  [6y + 1] - 4

7y - 2y + 14  =  6y + 6 - 4             Simplify each side:

       5y + 14  =  6y + 2                   Get all the y-terms to one side by subtracting 5y from both sides:

  5y - 5y + 14  =  6y - 5y + 2          Simplify each side:

                14  =  y + 2                   Subtract 2 from both sides:

          14 - 2  =  y + 2 - 2

                12  =  y

Check the answer:  7(12) - 2(12 - 7)  =  7(12) - 2(5)  =  84 - 10  =  74

                             6(12 + 1) - 4  =  6(13) - 4  =  78 - 4  =  74                         It checks!

7x² - 36x + 12

Since the last term is positive, the signs must be both positive or both negative.

Since the middle term is negative, the signs must be both negative.

There is only one set of factors for 7:  1 and 7.

However, there are many factors for 12:  12 and 1,  6 and 2,  4 and 3 (and they have to be checked both ways).

So the possibilities are:  

     (7x - 12)(x - 1)           (7x - 1)(x - 12)           (7x - 6)(x - 2)          (7x - 2)(x - 6)

     (7x - 4)(x - 3)             (7x - 3)(x - 4)

Check each of the above to find the proper factors.

If none of them work, then the problem cannot be factored using whole numbers.

A 1/2 exponent is a square root; 

so, 422 ^ (2)  means  √422  =  20.5426 ...

This is a perfect square trinomial:  4y^4 = (2y²)²   and   9x^4 = (3x²)²

Since the last sign is a '+' and the first term is a '-', it will factor into two factors, each one containing a minus sign.

Try   (2y² - 3x²)(2y² - 3x²) 

Since it has uniform acceleration: divide 25m/s by 5s to get 5 m/s/s or 5 m/s².

On a graph with the x-axis representing time and the y-axis representing velocity, the slope of the line represents acceleration.

The equation y = 5x will contain the two points (0,0) representing being stopped at time = 0s and the point (5,25) withich represents 25m/s at time = 5s.

The area under the curve will represent distance because the area = .5 · length · width. The figure with a constant slope from (0,0) to (5,0) to (5,25) is a triangle. The area represents distance because the length is seconds and the height is meters per second and when you multiply seconds x meters/second, you get meters.

So, the distance travelled is the area of a triangle whole base is 5 units and whole height is 25 units: 

.5 x 5 x 25  =  62.5 meters.

Try this:  (4y - x)(4y-x)

(3x - 4) + (1 - 2x)i  =  5 + 0i

Set the real part equal to the real part:                    3x - 4  =  5   --->   x = 3

Set the imaginary part equal to the imaginary part:   1 - 2x  = 0   --->   x = 1/2

Impossible!

7/9 + 11/13

Find a common denominator:  multiply  9 x 13  to get  117

To change 7/9:   (7 x 13) / (9 x 13)  =  91/117

to change 11/13:  (11 x 9) / (13 x 9)  =  99/117

91/117 + 99/117  = 190/117                       (Add the numerators 91 + 99; keep the denominator as it is.)

You have two variables; you can't solve to get a single answer.

However, you could solve for y, for x, for y/x, or x/y.

If you can be more specific, we can help you.

Make certain that you write both fractions with the same denominator.    

For instance, if the problem is  4 3/4  +  9 2/5:

Find the common denominator (multiplying them will give you a common denominator) 4 x 5 = 20

To change 3/4:  3/4 x 5/5  =  15/10   (multiply the numerators together: 3 x 5 = 15                                                                                                        and the denominators together: 2 x 5 = 10) 

To change 2/5:  2/5 x 4/4  =  8/20

So the problem becomes:  

    4 15/10  + 9 8/20   =   13 23/20     (Add the numerators together; put that answer over the common                                                                     denominator.)

23/20   =   1 3/20     so     13 23/20   =  13 + 1 3/20     =  14 3/20

I hope this helps. If you have specific questions, please re-pst.