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Travka [436]
3 years ago
6

Can a quadratic have one real and one non real root

Mathematics
1 answer:
alukav5142 [94]3 years ago
3 0

Answer:

Yes, it is possible. We already know that complex roots exist in conjugate pair but the case is true only when the equation has real coefficients. But on the other hand, if the equation has complex coefficients, then it is possible to attain one real and one complex root.

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Monthly sales of an SUV model are expected to increase at the rate of S′1t2 = -24t 1>3 SUVs per month, where t is time in mon
uysha [10]

The company will continue to manufacture this model for 19 months.

The question is ill-formatted. The understandable format is given below.

An SUV model's monthly sales are anticipated to rise at a rate of

S'(t)=-24^{1/3} SUVs each month, where "t" denotes the number of months and "S(t)" denotes the monthly sales of SUVs. When monthly sales of 300 SUVs are reached, the business intends to discontinue producing this model.

Find S if monthly sales of SUVs are 1,200 at time t=0 (t). How long will the business keep making this model?

Given S'(t)=-24^{1/3} ... ... (1)

Condition (1) at t=0; s(t) =1200

We find S(t)=300 then t=?

a) We find S(t)

from S'(t)=-24^{1/3}

\implies \frac{dS(t)}{dt}=-24t^{1/3} ~~[\because X'(t)=\frac{dX}{dt} \\\implies dS(t) = -24t^{1/3}dt\\

On Integrating both sides

S(t)=-18t^{4/3}+c ~...~...~(2)

Now, at t=0 then S(t)=1200

So, from (2)

1200=0+c

⇒c=1200

\thereforefrom eq (2)

S(t)=-18t^{4/3}+1200 ~...~...~(3)

b) Considering that the firm intends to cease production of this model once monthly sales exceed 300 SUVs.

So, take S(t)=300

from eq (2) 300=-18t^{4/3}+1200

t^{4/3}=\frac{1200-300}{18}\\=\frac{900}{18}=50\\\implies t=50^{3/4}\\\implies t=18.8030

Hence the company will continue 18.8020 months.

Learn more about integration here-

brainly.com/question/18125359

#SPJ10

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