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lesya [120]
3 years ago
7

There are 278 houses in Hannah's neighborhood. She collected a total of $780 from her neighbors to donate for a local charity. H

annah received $5 at each house she went to. From how many houses did Hannah NOT collect money from? Explain the answer and show your work
Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0
1. 780 / 5 = 156
2. 278 - 156 = 122

She did not collect any money from 122 houses.
ser-zykov [4K]3 years ago
6 0
First, we need to find out how many houses that Hannah received money from. We can do this by dividing the total amount of money she received by the amount she received per house.

780 / 5 = 156

Hannah received money from 156 houses.
To find the amount of houses she did NOT collect money from, we just need to subtract the amount of houses that she DID receive money from away from the total amount of houses in her neighborhood.

278 - 156 = 122

Hannah did not collect money from 122 houses.
Hope this helped! =)
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If f(x) = 9x10 tan−1x, find f '(x).
djverab [1.8K]

Answer:

\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           \displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:                                                                             \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = 9x^{10} \tan^{-1}(x)

<u>Step 2: Differentiate</u>

  1. [Function] Derivative Rule [Product Rule]:                                                   \displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
  2. Rewrite [Derivative Property - Multiplied Constant]:                                  \displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
  3. Basic Power Rule:                                                                                         \displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]
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2 years ago
B) The price of an electric fan is fixed 20% above its cost price. When it is sold allowing
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Answer:

\boxed{ \boxed{ \sf {Marked \: price \:  = Rs \: 1500}}}

\boxed{ \bold{ \boxed{ \sf{Selling \: price =  \: Rs \: 1230}}}}

Step-by-step explanation:

Let Cost price ( C.P ) be x

<u>Finding </u><u>the </u><u>Marked </u><u>price </u><u>and </u><u>selling </u><u>price </u>

Marked price = \sf{x + 20\% \: of \: x}

⇒\sf{x +  \frac{20}{100}  \times x}

⇒\sf{ \frac{x \times 100 + 20x}{100} }

⇒\sf{ \frac{120x}{100} } ⇒ ( i )

Selling price = \sf{marked \: price \:  - 18\% \: of \: marked \: price}

⇒\sf{ \frac{120x}{100} -  \frac{18}{100}  \times  \frac{120x}{100} }

⇒\sf{ \frac{120x}{100}  -  \frac{54x}{250} }

⇒\sf{ \frac{120x \times 5 - 54x \times 2}{500} }

⇒\sf{ \frac{600x - 108x}{500} }

⇒\sf{ \frac{492x}{500} } ⇒ ( ii )

<u>Finding </u><u>the </u><u>value </u><u>of </u><u>x </u><u>(</u><u> </u><u>Cost </u><u>price </u><u>)</u>

\sf{loss = cost \: price - selling \: price}

⇒\sf{20 = x -  \frac{492x}{500} }

⇒\sf{20  =  \frac{x \times 500 - 492x}{500} }

⇒\sf{20 =  \frac{8x}{500} }

⇒\sf{8x = 10000}

⇒\sf{x =  \frac{10000}{8} }

⇒\sf{x = \: Rs \:  1250}

Value of x ( cost price ) = Rs 1250

<u>Now</u><u>,</u><u> </u><u>Replacing </u><u>the </u><u>value </u><u>of </u><u>x </u><u>in </u><u>(</u><u> </u><u>i </u><u>)</u><u> </u><u>in </u><u>order </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>marked </u><u>price</u>

\sf{marked \: price =  \frac{120x}{100} }

⇒\sf{ \frac{120 \times 1250}{100} }

⇒\sf{   \: Rs \: 1500}

<u>Replacing </u><u>value </u><u>of </u><u>x </u><u>in </u><u>(</u><u> </u><u>ii </u><u>)</u><u> </u><u>in </u><u>order </u><u>to </u><u>find </u><u>the </u><u>value </u><u>of </u><u>selling </u><u>price</u>

\sf{selling \: price =  \frac{492 \: x}{500} }

⇒\sf{ \frac{615000}{500} }

⇒\sf{ \: Rs \: 1230}

Thus , Marked price of the fan = Rs 1500

Selling price of the fan = Rs 1230

Hope I helped!

Best regards!!

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