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nalin [4]
3 years ago
10

MY BRAIN IS DEAD! PLS HELP MEH ITS THE LAST ONE! TYSM! Time taken for 6 people to build a house is 30 days. How many days will i

t take for 2 people to build the same house ? (You need to think if this is a direct or inverse proportion problem).
Mathematics
2 answers:
iren2701 [21]3 years ago
8 0

Answer:

2 people will take 90 days to complete the same house

Step-by-step explanation:

<u><em>Given:</em></u>

T1 = 30 days

P1 = 6

T2 = x

P2 = 2

This is an inverse proportion because if there are more people, less time will be taken to complete the house and vice versa.

<u><em>So, We'll write it in the order of :</em></u>

=> T1:T2 = P2:P1

=> 30:x = 2:6

<u><em>Writing in fraction form</em></u>

=> \frac{30}{x} = \frac{2}{6}

<em>Cross Multiplying</em>

=> 2x = 180

<em>Dividing both sides by 2</em>

=> <u><em>x = 90 days</em></u>

So, <u><em>2 people will take 90 days to complete the same house</em></u>

PilotLPTM [1.2K]3 years ago
5 0

Answer:

90 days

Step-by-step explanation:

6 people- 30 days

1 person- 30*6= 180 days

2 people- 180/2= 90 days

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Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

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