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2 years ago

+find the discriminant and the number of real roots for this equation 9x^2+12x+14

Mathematics

2 answers:

Juli2301 [7.4K]2 years ago

8 0

Answer:

no real roots

Step-by-step explanation:

9x²+12x+14

recall that for a quadratic equation ax²+bx+c, the discriminant = b²- 4ac

In our case,

discriminant,

= b²- 4ac

= 12² - 4(9)(14)

= 144 - 504

= -360

because the discriminant is negative, there are no real roots.

Dimas [21]2 years ago

3 0

Answer:

the number of real roots for this equation 9x^2+12x+14 is :

0 (zero)

Step-by-step explanation:

Let Δ represent the discriminant

Δ = b² - 4ac = 12² - 4(9)(14) = -360

since Δ < 0 then it doesn’t have any real root.

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A production line has two machines, Machine A and Machine B, that are arranged in series. Each jol needs to processed by Machine

marysya [2.9K]

Answer:

a. Utilization of machine A = 0.8

Utilization of machine B = $\frac{2}{9}$

b. Throughput of the production system:

$E_S = \frac{E_A+E_B}{2} = \frac{20+\frac{18}{7} }{2}=(\frac{1}{2}*20 )+ (\frac{1}{2}*\frac{18}{7} )= 10+\frac{9}{7}= \frac{79}{7} mins$

c. Average waiting time at machine A = 16 mins

d. Long run average number of jobs for the entire production line = 3.375 jobs

e. Throughput of the production system when inter arrival time is 1 = $\frac{5}{6} mins$

Step-by-step explanation:

Machines A and B in the production line are arranged in series

Processing times for machines A and B are calculated thus;

$M_A = \frac{1}{4}/min$

$M_B = \frac{1}{2} /min$

Inter arrival time is given as 5 mins

$\beta _A = \frac{1}{5} = 0.2/min$

since the processing time for machine B adds up the processing time for machine A and the inter arrival time,

Inter arrival time for machine B,

$5+4 = 9mins\\\beta _B = \frac{1}{9} /min$

a. Utilization can be defined as the proportion of time when a machine is in use, and is given by the formula $\frac{\beta }{M}$

Therefore the utilization of machine A is,

$P_A = \frac{\beta_A }{M_A}=\frac{0.2}{\frac{1}{4} }= 0.8$

And utilization of machine B is,

$P_B = \frac{\beta_B }{M_B} = \frac{\frac{1}{9} }{\frac{1}{2} }= \frac{2}{9}$

b. Throughput can be defined as the number of jobs performed in a system per unit time.

Throughput of machines A and B,

$E_A = \frac{\frac{1}{M_A} }{1-P_A}= \frac{4}{1-0.8} = \frac{4}{0.2}= 20 mins\\ E_B = \frac{\frac{1}{M_B} }{1-P_B}= \frac{2}{1-\frac{2}{9} } = \frac{18}{7}mins$

Throughput of the production system is therefore the mean throughput,

$E_S = \frac{E_A+E_B}{2} = \frac{20+\frac{18}{7} }{2}=(\frac{1}{2}*20 )+ (\frac{1}{2}*\frac{18}{7} )= 10+\frac{9}{7}= \frac{79}{7} mins$

c. Average waiting time according to Little's law is defined as the average queue length divided by the average arrival rate

Average queue length, $L_q = \frac{P_A^2}{1-P_A} = \frac{0.8^2}{1-0.8}=\frac{0.64}{0.2}= 3.2$

Average waiting time = $\frac{3.2}{\frac{1}{5} }= 3.2*5=16mins$

d. Since the average production time per job is 30 mins;

Probability when machine A completes in 30 mins,

$P(A = 30)= e^{-M_A(1-P_A)30 }= e^{-\frac{1}{4}(1-0.8)30 }=0.225$

And probability when machine B completes in 30 mins,

$P(B = 30)= e^{-M_B(1-P_B)30 }= e^{-0.5(1-\frac{2}{9} )30 }=e^{-\frac{15*7}{9} }=e^{-11.6}$

The long run average number of jobs in the entire production line can be found thus;

$P(S = 30)=(\frac{ {P_A}+{P_B}}{2})*30 = (\frac{ 0.225}+{0}}{2})*30= 0.1125*30\\=3.375jobs$

e. If the mean inter arrival time is changed to 1 minute

$\beta _A= \frac{1}{1}= 1/min\\\beta _B= \frac{1}{6}/min\\ M_A = \frac{1}{4}min\\ M_B = \frac{1}{2} min$

Utilization of machine A, $P_A = \frac{\beta_A }{M_A} = 4$

Utilization of machine B, $P_B = \frac{\beta_B}{M_B} = \frac{1}{3}$

Throughput;

$E_A = \frac{\frac{1}{M_A} }{1-P_A} = \frac{4}{1-4} = \frac{4}{3} \\E_B= \frac{\frac{1}{M_B} }{1-P_B} = \frac{2}{1-\frac{1}{3} } = 3\\\\E_S= \frac{E_A+E_B}{2} = \frac{\frac{4}{3}+3 }{2}=(\frac{4}{3} *\frac{1}{2} )+(3*\frac{1}{2} ) =\frac{2}{3} + \frac{3}{2} \\= \frac{5}{6} min$