The cost price of the table is $40.
<h3>What is Gain ?</h3>
Gain is the amount gain by selling the product at a higher price than its cost.
Let the cost of the table is $ x
The percentage gain is x% (as given in the question)
Cost price = ?
It is known that
Step 1 : Gain = ( selling Price - Cost Price) * 100 / Cost Price
Selling price = 56
Cost Price = $ x
Therefore substituting the value
x = (56 - x) * 100 / x
x² = 5600 - 100x
x² +100x -5600 = 0
Step 2 : Factorizing
x² + 140x - 40 x -5600 = 0
x( x+14 ) -40( x +14) = 0
( x - 40)(x +14) = 0
x = $40
Therefore the cost price of the table is $40.
To know more about Gain
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I = $ 1,200,000.00
Equation:
I = Prt
Calculation:
First, converting R percent to r a decimal
r = R/100 = 3%/100 = 0.03 per year,
then, solving our equation
I = 1000000 × 0.03 × 40 = 1200000
I = $ 1,200,000.00
The simple interest accumulated
on a principal of $ 1,000,000.00
at a rate of 3% per year
for 40 years is $ 1,200,000.00.
Answer:
espero que esto ayude
Step-by-step explanation:
length = int (input ("Ingresa una longitud"))
width = int (input ("Ingresa un ancho"))
área = largo * ancho
imprimir ("El área es" + área)
Answer:
it let me now!!
Step-by-step explanation:
so i cant put an answer for some reason but here it is: when dividing fractions you would need to multiply to solve. The easyist way to do so from the ways i learned is keep change flip. so keep: 2/3 change a divide to a multiplication simbol and flip 1/4 to 4/1 so your left with G: 2/3 x 4/1
hope that helps btw if you need me to explain more let me know whats confusing you so i can help!!
Answer:
They sold 60 muffins and 60 cookies yesterday
It is a LCM problem
Step-by-step explanation:
6 muffins together
20 cookies together
To sell the same amount of muffins and cookies,
Find the lowest common multiple of 6 muffins and 20 cookies
6 muffins = 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78
20 cookies = 40, 60, 80
The lowest common multiple of 6 muffins and 20 cookies is 60
Therefore,
They sold 60 muffins and 60 cookies yesterday
It is a LCM problem
