Ask question

Ask question

All categories

3 years ago

5

a television regulary cost $180 but it is on sale for 40% off this is a 5% sales tax on the sale price how much does the televei

sion cost after the discount and sales tax

1 answer:

nadya68 [22]3 years ago

8 0

If it is 40% off then it is 100-40 = 60% of original price. So we do 180$ * . 6 = $108 before tax. To add on tax we do: 100 + 5 = 105% of that price.

You might be interested in

Work out the percentage change to 2 decimal places when a price of £57.99 is decreased to £49.99.

IRISSAK [1]

Answer:

Percentage change in price = 13.80%

Step-by-step explanation:

Given that:

Original price = £57.99

Decreased price = £49.99

Difference = Original price - Decreased price

Difference = 57.99 - 49.99

Difference = £8.00

Percent change = $\frac{Difference}{Original\ price}100$

Percent change = $\frac{8.00}{57.99}*100$

Percent change in price = 13.80%

Hence,

Percentage change = 13.80%

5 0

3 years ago

1300 error interval to 2dp

Degger [83]

Answer:

Huh

Step-by-step explanation:

5 0

3 years ago

Samuel pays $20 for a shirt that is on sale. If he saved $5, what is the percent of the discount?

mamaluj [8]

Hi Kyliebreland!
So, If he saved $5 then we do $20 divided by $5 which gives 4 so now we split 100% into 4 pieces which gives us 25%.
Hope this helps!

7 0

4 years ago

monitta

Answer:

D. y - 3 = -(x + 5)

Step-by-step explanation:

I used a calculator called m.a.t.h.w.a.y !

7 0

3 years ago

Read 2 more answers

What is the value of \dfrac{d}{dx}\left(\dfrac{2x+3}{3x^2-4}\right) dx d ( 3x 2 −4 2x+3 )start fraction, d, divided by, d, x

stellarik [79]

Answer:

4.

Step-by-step explanation:

We are asked to find the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ .

First of all, we will find the derivative of the given expression using "Quotient Rule of Derivatives" as shown below:

$(\frac{f(x)}{g(x)})'=\frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{(g(x))^2}$

$\frac{d}{dx}(\frac{2x+3}{3x^2-4})$

$\frac{\frac{d}{dx}(2x+3)*(3x^2-4)-(2x+3)*\frac{d}{dx}(3x^2-4)}{(3x^2-4)^2}$

$\frac{2*(3x^2-4)-(2x+3)*(6x)}{(3x^2-4)^2}$

$\frac{6x^2-8-12x^2-18x}{(3x^2-4)^2}$

$\frac{-6x^2-18x-8}{(3x^2-4)^2}$

Therefore, our required derivative is $\frac{-6x^2-18x-8}{(3x^2-4)^2}$ .

Now, we will substitute $x=-1$ in our derivative to find the required value as:

$\frac{-6(-1)^2-18(-1)-8}{(3(-1)^2-4)^2}$

$\frac{-6(1)+18-8}{(3(1)-4)^2}$

$\frac{-6+18-8}{(3-4)^2}$

$\frac{4}{(-1)^2}$

$\frac{4}{1}$

$4$

Therefore, the value of expression $\frac{d}{dx}(\frac{2x+3}{3x^2-4})$ at $x=-1$ is 4.

6 0

3 years ago