HCF of algebraic expression x square minus y square and x y minus y square. ​

(x-y+z)(x-y-z)

Given (x²-2xy+y²) - z²

= (x-y)² - z²

/* By algebraic identity

a²-2ab+b² = (a-b)² */

=[(x-y)+z][(x-y)-z]

/* a²-b² = (a+b)(a-b) */

= (x-y+z)(x-y-z)

Therefore,

(x²-2xy+y²)-z² =(x-y+z)(x-y-z)

Step-by-step explanation:

x

2

y

2

=x×x×y×y

xy

2

=x×y×y

h.c.f=xy

2

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation: