Q2 (i). A line “t” is parallel to 3y = 6x + 9. Find the slope of this line “t”. (ii) Another line “r” is perpendicular to the line 3y = 6x + 9. Find the gradient of the line “r”. plz can anyone tell me by doing the practice on the copy I will be thankful

1) 2

2) -1/2

Step-by-step explanation:

1) 3y= 6x+9

y= 2x+3

Slope is 2

Parallel line "t" has the same slope, it will have equation:

y= 2x+b

2) y =2x+3

Perpendicular line"r" has a slope opposite-reciprocal to this, so the slope will be -1/2, the equation for line"r" is:

y= -1/2x +b

The gradient of the line is same as slope and it is -1/2 for line"r"

step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

from the figure we get,

the x axis shows   the months and,

the y-axis shows the number of satisfied customer.

each line or each cell in upward direction (y-axis) shows 500.

the appropriate scale for the y - axis is: -

a = 2000 customers

b = 4000 customers

c = 6000 customers

d = 8000 customers

e = 10000 customers

f = 12000 customers

g = 14000 customers