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n200080 [17]
3 years ago
15

1/y - 1/x = 1/60 3y - 2x = 6 (Simultaneous question)

Mathematics
1 answer:
Daniel [21]3 years ago
6 0

1y - 1/x = 1/60 = x*y =60 = x=60/y

3y - 2(60/y) = 6

3y^2-120=6y

3(y^2-40-2y) =0

y^2-40-2y=0

y = 7.4


3(7.4)-2x=6

22.2-2x = 6

2x = -16.2

x = -8.1



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Answer:

x = 1

Step-by-step explanation:

12.3 + 2.3 = 14.6

If you multiplied 12.3 by 1 it would still be 12.3

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the flower shop has 40 times as many flowers in one cooler as Julia has in her bouquet. The cooler has 120 flowers. How many flo
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3 years ago
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Suppose a, b denotes of the quadratic polynomial x² + 20x - 2022 & c, d are roots of x² - 20x + 2022 then the value of ac(a
Alja [10]
<h3><u>Correct Question :- </u></h3>

\sf\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0 \: and \:  \\  \sf \: c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0 \: then \:

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) =

(a) 0

(b) 8000

(c) 8080

(d) 16000

\large\underline{\sf{Solution-}}

Given that

\red{\rm :\longmapsto\:a,b \: are \: the \: roots \: of \:  {x}^{2} + 20x - 2020 = 0}

We know

\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}

\rm \implies\:ab = \dfrac{ - 2020}{1}  =  - 2020

And

\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}

\rm \implies\:a + b = -  \dfrac{20}{1}  =  - 20

Also, given that

\red{\rm :\longmapsto\:c,d \: are \: the \: roots \: of \:  {x}^{2}  -  20x  + 2020 = 0}

\rm \implies\:c + d = -  \dfrac{( - 20)}{1}  =  20

and

\rm \implies\:cd = \dfrac{2020}{1}  = 2020

Now, Consider

\sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d)

\sf \:  =  {ca}^{2} -  {ac}^{2} +  {da}^{2} -  {ad}^{2} +  {cb}^{2} -  {bc}^{2} +  {db}^{2} -  {bd}^{2}

\sf \:  =  {a}^{2}(c + d) +  {b}^{2}(c + d) -  {c}^{2}(a + b) -  {d}^{2}(a + b)

\sf \:  = (c + d)( {a}^{2} +  {b}^{2}) - (a + b)( {c}^{2} +  {d}^{2})

\sf \:  = 20( {a}^{2} +  {b}^{2}) + 20( {c}^{2} +  {d}^{2})

\sf \:  = 20\bigg[ {a}^{2} +  {b}^{2} + {c}^{2} +  {d}^{2}\bigg]

We know,

\boxed{\tt{  { \alpha }^{2}  +  { \beta }^{2}  =  {( \alpha   + \beta) }^{2}  - 2 \alpha  \beta  \: }}

So, using this, we get

\sf \:  = 20\bigg[ {(a + b)}^{2} - 2ab +  {(c + d)}^{2} - 2cd\bigg]

\sf \:  = 20\bigg[ {( - 20)}^{2} +  2(2020) +  {(20)}^{2} - 2(2020)\bigg]

\sf \:  = 20\bigg[ 400 + 400\bigg]

\sf \:  = 20\bigg[ 800\bigg]

\sf \:  = 16000

Hence,

\boxed{\tt{ \sf \: ac(a - c) + ad(a - d) + bc(b - c) + bd(b - d) = 16000}}

<em>So, option (d) is correct.</em>

4 0
2 years ago
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Answer:

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d. 34.56 rounded to the nearest tenth is 34.6

This makes it so that C. 34.49 is your answer.

~

4 0
3 years ago
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