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3 years ago

14

Tony plans to run a 5,000-meter fun run at a constant rate of 250 meters per minute. He uses function f to model his distance fr

om the finish line x minutes after the start of the race.
x f(x)
0 5,000
1 4,750
2 4,500
3 4,250
4 4,000
5 3,750
6 3,500

A.
Tony's function is always decreasing.
B.
Tony's function is always negative.
C.
Tony's function is exponential.
D.
The domain of Tony's function is [0 , 5,000].

2 answers:

kramer3 years ago

8 0

Answer:

Option A.

Step-by-step explanation:

Based on the information provided on the question

the function used to model Tony's run is:

f(x) = -250*x + 5000

This means that after x = 20 minutes, Tony will arrive to the finish line

f(20) = -250*(20) + 5000 = -5000 + 5000 = 0

The function is always decreasing, because we are dealing with a line with negative slope.

Option A.

Elden [556K]3 years ago

5 0

Answer:

Step-by-step explanation:

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3 years ago

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