1. the answer is 1.

If Y and y are the same, then it is -y + 4 or 4 - y.

Y - 2x +4 is an expression which cannot be simplified. Since it is neither an equation nor an inequality, it cannot be solved.

(y^2 - 4) / (y + 2) = [(y -2)(y+2)]/(y+2) = y-2

y-2 +2 y=2 slope is undefined y intercept is 2

When Y = 8, Y - 2 = 6.

There is no answer to y - 2, because y - 2 isn't a question.

is that (y*2)? if yes the answer will be:

2-6y^2

Y/2

If you mean y/2 + y (without parentheses), y/2 is the same as (1/2)y, and y is the same as 1y. You can use the distributive property to combine this; in this case, the result is (3/2)y.

y = 3x - 2

3x = y + 2

x = y + 2

---------

3

If you mean: (y-2)(y+2) then it is y^2 -4 which is the difference of two squares

x/y = 5/(y^2) and y/x = 5/(x^2). So x/y + y/x = 5/(y^2) + 5/(x^2), which equals 5x^2/(x^2 y^2) + 5y^2/(x^2 y^2) equals 5(x^2 + y^2)/25 equals (x^2 + y^2)/5. x^2 = 25/y^2, so you get (25/y^2 + y^4/y^2)/5 equals ((25 + y^4)/y^2)/25, which shows that your math teacher is on crack. Seriously, I'm not sure that's true.

The expression y^4 - z^4 can be factored using the difference of squares formula. It can be written as (y^2 + z^2)(y^2 - z^2), which can be further simplified to (y^2 + z^2)(y + z)(y - z).

y3 + 2y2 - 81y -162 = y2(y+2) - 81(y+2) =(y+2)(y2-81) = (y+2)(y-9)(y+9)

6

2

y = 15 so that 13= 15 - 2

y-2 = 13

y-2 +2 = 13 +2

y = 15

X - Y = 2

When X = 3 you have 3 - Y = 2

Moving Y to RHS: 3 = Y + 2

Moving 2 to LHS: 1 = Y

y^2 X y^3 = y^(2 + 3) = y^5

You can only do this if the coefficient 'y' is the same for both terms.

Remember

y^2 = y X y

y^3 = y X y X y

Hence

y^2 X y^3 = y X y X y X y X y = y^5

Similarly for division/subtraction

y^3 / y^2 = y^(3 - 2 ) = y^1 = y The power of '1' is trivial and not normally shown.

NB You CANNOT do z^2 X y^3 by adding the indices.

z^2 X y^3 is (z^2)*(y^3)

8y/y + 2 + 16/y + 2 = 8 + 2 + 16/y +2 = 12 + 16/y = 4(3 + 4/y) or 4/y*(3y + 4)

y(y - 2)(y + 2)

There is no answer to y - 2, because y - 2 isn't a question.

There is no answer to y - 2, because y - 2 isn't a question.

It is the expression y/2.

1 - y^(3)=

1^(3) - (3) =

This factors to

(1 - y)(1^(2) + y + y^(2))

More simply written as

( 1 - y)(1 + y + y^(2))

y+2x+2=4

y=-2x-2+4

y=-2x+2

To find the y-intercept, let x=0

y=2

The y-intercept is 2. Or, since you've written the equation of the line in the slope-intercept form, y = mx + b, where m is the slope, and b is the y-intercept, as y = -2x + 2, you can say that b = 2, so that y-intercept is 2.

If you mean y=3x-2, you can solve for x as follows:

3x-2+2 = y+2, x = (y+2)/3 = y/3 + 2/3

y2 + y2 - 56 let's factor it such as

= 2y2 - 56

= 2(y2 - 28)

= 2[y2 - (√28)2]

= 2(y - √28)(y + √28)

= 2(y - √4*7)(y + √4*7)

= 2(y - 2√7)(y + 2√7)

2.(2+y) = 2.2 + 2.y = 4 + 2y

Yes

if x=2 and y=2, then x=y

1

Any two numbers of this type can be solved using boolean logics.......

It is very simple once had a glance on it .

Here is a method to prove x=y where x and y are any numbers.....

let XY = XY

We can rewrite the above as

ð X2 - (X+Y)*X = Y2 - (X+Y)*Y

ð Multiplying both sides with 2/2 we get

ð X2 - 2*x*(X+Y)/2 = Y2 - 2*Y*(X+Y)/2

ð Adding (X+Y)2/2 on both sides we get

ð X2 - 2*x*(X+Y)/2 + (X+Y)2/2= Y2 - 2*Y*(X+Y)/2 + (X+Y)2/2

ð This is in the form of a2-2ab +b2 = (a+b)2

ð (X - (X+Y)/2)2 = (Y-(X+Y)/2)2

ð Canceling the squares on both sides(case condition)

ð X-(X+Y)/2 = Y - (X+Y)/2

ð Canceling (X+Y)/2 on both sides

ð X=Y

Only when X or Y = 0

That is because (x + y) squared = x^2 + 2xy + y^2

x^2 + 2xy + y^2 = x^2 + y^2 when x or y = 0

(-2)2 = +4

2(2+Y)

(y-7)(y-2) => y² - 2y - 7y + 14

Answer: y² - 9y + 14.

{2+(2x-y)-(3x-y)

Technically, no because a perimeter is a linear measure and an area is a square measure. However, there are infinitely many rectangles such that the NUMERICAL VALUE of their perimeter is the same as the NUMERICAL value of their area.

Select any number y greater than or equal to 4.

let x = 2*y/(y-2)

Consider the rectangle with dimensions width x and length y.

Its area is x*y = [2*y/(y-2)]*y = 2y2/(y-2) square units.

Its perimeter is 2(x+y) = 2*[(2y/(y-2) + y] = 2/(y-2)*[2y+y*(y-2)]

= 2/(y-2)*[2y+y2-2y] = 2/(y-2)*y2 = 2y2/(y-2) units

Since y is an arbitrary number greater than 4, there are infinitely many choices for y giving rise to infinitely many shapes. The one with the smallest y: y = 4 is actually a square - with sides of 4 units and perimeter/area = 16.

2 + y = 16 - y

Add y to both sides: 2 + 2y = 16

Subtract 2 from both sides: 2y = 14

Divide both sides by 2: y = 7

-2(y plus 5) plus y?

If Y=5x + 2 and x is 5 then

Y= 5*5 +2, using order of operations, you get

Y=25 +2

Y= 27

y = 2 is a horizontal line that crosses the y axis at 2.

examples of linear relationships are:

- Y= mX + b

-Y= mx

-Y= 2(6) + 2

- Y= 2(6) - 2

- Y= 3(5)

y = x + 2

y = -x + 4

x + 2 = -x + 4

2x + 2 = 4

2x = 2

x = 1

y = x + 2

y = 1 + 2

y = 3

(1, 3)

y=3x-2 has gradient or slope of 3 and y intercept of -2 y=3x+2 has the same slope or gradient but y intercept of 2 in general, y=mx+b has a slope of m and a y intercept of b

4y2 + 16y + 16 = 4(y2 + 4y + 4) = 4(y2 + 2y + 2y + 2) = 4[y(y + 2) + 2(y + 2)]

= 4(y + 2)2

If that should be y = 2 the slope is 0 and the y intercept is 2.