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Margaret [11]
3 years ago
13

(b) How much the selling price should be fixed for pulse bought for Rs.70 per kg. to earn a profit of Rs.6 after allowing a 5 %

discount?
​
Mathematics
1 answer:
Mama L [17]3 years ago
3 0

Answer:

Rs. 80

Step-by-step explanation:

Given that :

Purchase price = 70

Profit = 6

Discount = 5%

Let selling price = x

Selling price * (1 - discount) = (purchase price + profit)

x * (1 - 5%) = (70 + 6)

x * (1 - 0.05) = 76

x * 0.95 = 76

0.95x = 76

x = 76 / 0.95

x = 80

Hence, selling price = Rs. 80

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Evgesh-ka [11]

Answer:

325108

Step-by-step explanation:

it is written as the above

7 0
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What is the degree of each polynomial?
Lana71 [14]

(1) Degree 4  (x^4)

(2) Degree 7 (x^7)

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Read 2 more answers
Please show me how to do this.
Aloiza [94]

Answer:

f(x) = \dfrac{1}{2}(x - \ln (x - 1) - 2)

Step-by-step explanation:

f'(x) = \dfrac{1}{2} - \dfrac{1}{2x - 2}

f(x) = \int (\dfrac{1}{2} - \dfrac{1}{2x - 2}) dx

f(x) = \int [\dfrac{1}{2} - \dfrac{1}{2(x - 2)}] dx

f(x) = \int \dfrac{1}{2}dx - \dfrac{1}{2} \int \dfrac{1}{x - 1} dx

f(x) = \dfrac{1}{2}x - \dfrac{1}{2} \ln (x - 1) + C

f(x) = \dfrac{x - \ln (x - 1)}{2} + C

f(2) = \dfrac{2 - \ln (2 - 1)}{2} + C = 0

f(2) = \dfrac{2 - \ln (1)}{2} + C = 0

f(2) = \dfrac{2}{2} + C = 0

f(2) = 1 + C = 0

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7 0
3 years ago
HELP PLZ!!! Michelle owns a clothing store that designs T-shirts and shorts. She sells
bixtya [17]
Answer: B

Step by Step: If it takes 2 hours to design shorts to get $15 and 1 hours to make shirts for $8 you can multiply 8 by 2 and get 16 which is more than 15. The reason you multiply 8 by 2 is because it takes 2 hours to make 1 pair of shorts. With this you then multiply 16 by 9 which gives you 144.

Hope this helps
3 0
3 years ago
If 5 + 3xy + 4y^3 = 0 then find dy/dx in terms of x and y
mafiozo [28]

Answer:

\displaystyle \frac{dy}{dx}=\frac{-3y}{3x+ 12y^2}

Step-by-step explanation:

<u>Implicit Differentiation</u>

We use implicit differentiation when it's not possible to find an expression of y as a function of x, or the expression is very hard to differentiate.

The implicit differentiation takes the original equation and differentiates each term, usually applying the product, quotient, power, or other similar rules.

In the course of the differentiation, we'll use f' as the derivative of f.

We'll find y'=dy/dx in the following equation:

5 + 3xy + 4y^3 = 0

Differentiating:

(5)' + (3xy)' + (4y^3)' = (0)'

The derivative of a constant is 0, thus:

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The first term is a product of variables, so we apply the product rule:

(f.g)'=f'.g+f.g'

The second term is the power of y. We apply the chain rule:

[f(g)]'=f'.g'

3(x'y+xy') + 4(3y^2y') = 0

Operating:

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Since x'=1:

3y+3xy'+ 12y^2y' = 0

Subtracting 3y:

3xy'+ 12y^2y' = -3y

Take y' as a common factor:

y'(3x+ 12y^2) = -3y

Solve for y':

\displaystyle y'=\frac{-3y}{3x+ 12y^2}

\boxed{\displaystyle \frac{dy}{dx}=\frac{-3y}{3x+ 12y^2}}

4 0
3 years ago
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