Harpreet gave her car for service at the service station and was charged as follows: a) 3.1 litres engine oil at the rate of rupees 210 per litre and VAT 20%. b) rupees 1105 for all other services and VAT 12.5%. c) rupees 1600 as labour charges and service tax 10%. find the bill amount.​

Bill amount = 3,784.325 rupees.

Step-by-step explanation:

Calculating each service charge rendered to Harpreet, we have

a) 3.1 litres engine oil at the rate of rupees 210 per litre and VAT 20%

= 3.1 L X 210 rupees=651 rupees

VAT 20 % = 651 X 20%=130.2 Rupees

Charge on purchase of engine oil = 651 rupees + 130 .2 rupees=781.2 rupees

b) All other services = 1105

VAT 12.5% = 1105 X 12.5 %=138.125 Rupees

Charge on all other services = 1,105+ 138.125 rupees=1,243.125 rupees.

c.Labour charges = 1600

Service tax 10% = 1600 x10%=160

Charge on Labour = 1600+160=1760 rupees.

Bill amount = Total Charge on all services rendered at the station =781.2 rupees + 1,243.125 rupees + 1760 rupees= 3,784.325 rupees.

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}