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Mathematics, 21.05.2020 03:12 thao6570

Please help me with this ASAP


Please help me with this ASAP

Answers

ansver
Answer from: Quest

5y

step-by-step explanation:

ansver
Answer from: Quest
I’m not gonna do that
ansver
Answer from: Quest

r(x)=-\frac{1}{50}x^2+14x

step-by-step explanation:

let's assume

number of candles set =x

price of of x candles =p(x)

we are given

a candle maker prices one set of scented candles at $10 and sells an average of 200 sets each week

so, we get point as (200,10)

x=200,p=10

he finds that when he reduces the price by $1, he then sells 50 more candle sets each week

so, we can find slope

m=-\frac{1}{50}

now, we can use point slope form of line

p(x)=mx+b

now, we can plug values

p(x)=-\frac{1}{50}x+b

we can use point

x=200,p=10

and find b

10=-\frac{1}{50}\times 200+b

10=-4+b

b=14

now, we can plug it back

and we get

p(x)=-\frac{1}{50}x+14

now, we can find revenue

we know that

revenue = price*quantity

so, we get

r(x)=x\times p(x)

r(x)=x\times (-\frac{1}{50}x+14)

r(x)=-\frac{1}{50}x^2+14x

ansver
Answer from: Quest

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.

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