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2 years ago

14

A dealer marks the price of his mobiles 40% above the cost price and allows 20% discount. If his/her purchase price of a mobile

is Rs 7,500, how much should a customer pay for it with 13% VAT?

1 answer:

Katena32 [7]2 years ago

5 0

Answer:

$8475

Step-by-step explanation:

13% of 7500 is 975 so i then added 975 7,500 and got $8,475

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a car rental company charges 10$ per day plus 1.00 per mile to rent a car. write an equation showing the total it would cost(y)

Alenkinab [10]

Answer:

I think this is what you are asking for hope it helps: cost=.17M+50

Step-by-step explanation:

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3 years ago

1 5 y + 8 − 4 5 = 9 10 y − 9 20

emmainna [20.7K]

Answer:

y ≈ .98 i hope this helps :)

Step-by-step explanation:

given equation 15y + 8 - 45 = 910y - 920

simplify the equation 15y - 37 = 910y - 920

subtract 15y from both sides -37 = 895y - 920

add 920 to both sides 883 = 895y

divide both sides by 895 .98 ≈ y

y ≈ .98

0.98659217877

7 0

3 years ago

Does anyone know which table of ordered pairs represents a proportional relationship!?

Law Incorporation [45]

It is table 3.

It’s the only one that has an equal division each time

(6/3) = 2
(10/5) = 2
(14/7) = 2

4 0

3 years ago

Read 2 more answers

Find cscx if sinx+cotx cosx= sqrt3

valentinak56 [21]

Answer:

The answer is (d) ⇒ cscx = √3

Step-by-step explanation:

∵ sinx + (cotx)(cosx) = √3

∵ sinx + (cosx/sinx)(cosx) = √3

∴ sinx + cos²x/sinx = √3

∵ cos²x = 1 - sin²x

∴ sinx + (1 - sin²x)/sinx = √3 ⇒ make L.C.M

∴ (sin²x + 1 - sin²x)/sinx = √3

∴ 1/sinx = √3

∵ 1/sinx = cscx

∴ cscx = √3

5 0

3 years ago

Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away. She is _____ per

valkas [14]

Answer:

Tiffany is 34% closer to the school.

Step-by-step explanation:

Given the statement: Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away.

She lived away from the school = 15 km

as, she moved 9.9 km away.

Now, the distance closer to school from where she lived = 15 -9.9 = 5.1 km

To, find how much percent she is closer to school.

Percent states that a number or ratio expressed as a fraction of 100.

$percent = \frac{5.1}{15} \times 100$ = $\frac{51 \times 100}{150} = \frac{51 \times 10}{15}$

Simplify:

percent closer to school = 34%

Therefore, she is 34% closer to the school.

5 0

3 years ago