All categories

3 years ago

The regular price of an item is 84$ the item is on sale with a discount of $12 what is the discount rate

Mathematics

1 answer:

Reika [66]3 years ago

5 0

12/84 = 1/7 ≈ 0.14
The discount rate was 14%

You might be interested in

if you a coin and roll a 6-sided die , what is the probability that you will flip a tails and roll a 2?

Aleonysh [2.5K]

1/2 for a coin. 1/6 for a die. multiply them and u get 1/12 (you happy now weirkai27?) 1/12 is your answer.

brainliest?

4 0

3 years ago

I need help!!

Natasha2012 [34]

Answer:

The exact solution is

$x = \frac{Ln(60)}{7*Ln(5)}$

And the approximation to three decimal places is:

x = 0.363

Step-by-step explanation:

Here we have the equation:

$5^{7*x} = 60\\$

Now we can remember a property of the natural logarithm function:

Ln(a^n) = n*Ln(a)

Now we can apply the Ln( ) function to both sides of that equation to get:

$Ln(5^{7*x}) = Ln(60)$

Then we get:

$7*x*Ln(5) = Ln(60)$

Solving that for x we get:

$x = \frac{Ln(60)}{7*Ln(5)} = 0.363$

So the exact solution is:

$x = \frac{Ln(60)}{7*Ln(5)}$

And the approximation to three decimal places is:

x = 0.363

7 0

3 years ago

0.51 + 0.63 = ?<br> and<br> 1.93 + 1.73 = ?

Anastasy [175]

For the first one: 1.14
Second one: 3.66

5 0

3 years ago

Read 2 more answers

PLEASE HELP I CANT FIGURE THEM OUT

statuscvo [17]

Answer:they all equal 69

Step-by-step explanation:

6 0

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago