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Alborosie
3 years ago
7

prakrit bought a pack of paper for 5.69 and printer toner for 9.76 he paid for with a 20 what was his change

Mathematics
2 answers:
Nady [450]3 years ago
7 0
You add 5.69 and 9.76 and you get 15.45. Then you subtract 20 from 15.45 and your get the change which is 4.55
natka813 [3]3 years ago
3 0
$4.55 Add 5.95 and 9.76. You should get 15.45. Then you do 20-15.45=4.55
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What is the sum of the roots of the quadratic equation x²+6x-14=0?​
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Answer:

<h3>Sum of Roots =<u> (-8)</u><u> </u><u> </u></h3><h3>Product of Roots = <u> </u><u> </u><u>(-84)</u><u> </u><u> </u></h3>

Step-by-step explanation:

{x}^{2}  + 6x - 14 = 0

Roots :- 6 and -14

Sum of the roots :

[6 + (-14)]

= (-8)//

Product of the roots :

[(6)(-14)]

= (-84)//

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Find the missing factor. 6x^2+7x-5=(3x+5)( )
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The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E
nexus9112 [7]

Answer:

The projected enrollment is \lim_{t \to \infty} E(t)=10,000

Step-by-step explanation:

Consider the provided projected rate.

E'(t) = 12000(t + 9)^{\frac{-3}{2}}

Integrate the above function.

E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt

E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c

2000=-\frac{24000}{3}+c

2000=-8000+c

c=10,000

Therefore, E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

Now we need to find \lim_{t \to \infty} E(t)

\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000

\lim_{t \to \infty} E(t)=10,000

Hence, the projected enrollment is \lim_{t \to \infty} E(t)=10,000

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