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mario62 [17]
2 years ago
5

A confectioner bought 15 chocolates for Rs. 1. If he sold all chocolates at 25% profit, how many chocolates were sold for Rs. 1?

​please step by step answer without and link or something please
Mathematics
1 answer:
shutvik [7]2 years ago
4 0

Answer:

He sold 12 chocolates for 1 Re.

Step-by-step explanation:

He bought 15 chocolates for 1 Rupee

He sold for a profit of 25%

Sale price = profit + cost price

SP= 25%of 1 + 1 Re

SP = 0.25*1+ 1

SP = 1.25 Rs

He sold 15 chocolates fro 1.25 Rs

Chocolates            Rupees

15                            1.25

x                                1

Applying cross product

x*1.25= 15*1

x= 15*1/1.25

x= 12

He sold 12 chocolates for 1 Re.

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +
Veronika [31]

Answer:

Verified

y(x) = \frac{3Ln(x) + 3}{x}

y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

                           x^2y' +xy = 3

- A general solution to the above ODE is also given as:

                          y = \frac{3Ln(x) + C  }{x}

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

                          y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x)  }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

                          -\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3

- Therefore, the complete solution to the given ODE can be expressed as:

                        y ( x ) = \frac{3Ln(x) + 3 }{x}

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

                         y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)

- Therefore, the complete solution to the given ODE can be expressed as:

                        y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}

                           

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6 0
3 years ago
Which expression represents the distance, in miles, between Paige's house and Diego's house?
lana [24]
Distance between <span>Paige's house and Diego's house 
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answer
</span>|3 - (-8)|
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3 years ago
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Answer:

15/60×100=25%

Step-by-step explanation:

the number of students wanted to go outside/the whole number of students×100

15/60×100=25%

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3 years ago
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Ulleksa [173]

Answer:

Step-by-step explanation:

Volumes of two spheres A and B = 648 cm³ and 1029 cm³

Things to remember:

1). Scale factor of two objects = \frac{r_1}{r_2} [r_1 and r_2 are the radii of two circles]

2). Area scale factor = \frac{(r_1)^2}{(r_2)^2}

3). Volume scale factor = \frac{(r_1)^3}{(r_2)^3}

Volume scale factor Or Volume ratio = \frac{V_A}{V_B}

                         \frac{(r_1)^3}{(r_2)^3}= \frac{648}{1029}

                         \frac{r_1}{r_2}=\sqrt[3]{\frac{648}{1029} }

                         \frac{r_1}{r_2}=\frac{6(\sqrt[3]{3})}{7(\sqrt[3]{3})}

                        \frac{r_1}{r_2}=\frac{6}{7}

Therefore, scale factor = \frac{r_1}{r_2}=\frac{6}{7}

                                      ≈ 6 : 7

Area scale factor Or area ratio = (\frac{r_1}{r_2})^2=(\frac{6}{7})^2

                                                   = \frac{36}{49}

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Volume scale factor or Volume ratio = \frac{648}{1029}

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                                                             ≈ 216 : 343

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mihalych1998 [28]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
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