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nikitadnepr [17]

3 years ago

6

The ratio of 5 rupee notes to 10 rupee notes in a purse is 3 : 4. There are 77 notes in the purse. How much money is there in th

e purse?

1 answer:

Norma-Jean [14]3 years ago

4 0

Answer:

605 rupees

Step-by-step explanation:

add the parts of the ratio 3 + 4 = 7

divide the number of notes by 7 to find one part of the ratio

$\frac{77}{7}$ = 11 ← 1 part of the ratio

3 parts = 3 × 11 = 33 ← number of 5 rupee notes

4 parts = 4 × 11 = 44 ← number of 10 rupee notes

Amount of money = (33 × 5 ) + (44 × 10) = 165 + 440 = 605 rupees

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Calculus 2. Please help

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<u>Calculus</u>

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Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

<u>Step 2: Integrate Pt. 1</u>

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

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<em>Rewrite u-substitution to format u-solve.</em>

Rewrite <em>du</em>: $\displaystyle dx = \frac{-1}{2x} \ dx$

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[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
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