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Marrrta [24]
3 years ago
5

Write the absolute value equation if it has the following solutions. Hint: Your equation should be written as |x−b| =c. (Here b

and c are some numbers.) Chapter Reference b Two solutions: x=2, x=12.
Mathematics
1 answer:
Over [174]3 years ago
6 0
<h2>Answer</h2>

|x-7|=5

<h2>Explanation</h2>

We know that we need to write our absolute value equation as |x-b|=c. We also know that the solutions must be x=2 and x=12. We need to replace those values for x in our absolute value equation, so we can create a system of equations and find the values of  b and c.

- For x=2

|x-b|=c

|2-b|=c equation (1)

- For x=12

|x-b|=c

|12-b|=c equation (2)

Now we can solve our system of equations step-by-step:

Step 1. Replace equation (1) in equation (2)

|12-b|=|2-b|

Step 2. Square both sides of the equation to get rid of the absolute values

|12-b|=|2-b|

(12-b)^2=(2-b)^2

Step 3. Use the square of a binomial formula: (a-b)^2=a^2-2ab+b^2 and solve the equation.

For our first binomial, (12-b)^2, a=12 and b=b; for our second binomial, (2-b)^2, a=2 and b=b

(12-b)^2=(2-b)^2

12^2-(2)(12)(b)+b^2=2^2-(2)(2)(b)+b^2

144-24b+b^2=4-4b+b^2

144-24b=4-4b

140=20b

b=\frac{140}{20}

b=7 equation (3)

Step 4. Replace equation (3) in equation (2) to find the value of c

|12-b|=c

|12-5|=c

|7|=c

c=7

Putting it all together we can conclude that our absolute value equation is |x-7|=5

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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

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3 years ago
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Answer: 3x=14

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Step-by-step explanation:

5x-y-2x+y=6+8

3x=14

X=14/3

Y=8+2x

Y=8+28/3

Y=52/3

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Answer:

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Suppose that X is a subset of Y. Let p be the proposition ‘x is an element ofX’ and let q be the proposition ‘x is an element of
Genrish500 [490]

Answer:

See answer below

Step-by-step explanation:

The statement ‘x is an element of Y \X’ means, by definition of set difference, that "x is and element of Y and x is not an element of X", WIth the propositions given, we can rewrite this as "p∧¬q". Let us prove the identities given using the definitions of intersection, union, difference and complement. We will prove them by showing that the sets in both sides of the equation have the same elements.

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ii) (I will abbreviate "if and only if" as "iff")

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iii) x∈A\(B U C) iff x∈A and x∉B∪C iff x∈A and x∉B and x∉C (if x∈B or x∈C then x∈B∪C thus we cannot have any of those two options). iff x∈A and x∉B and x∈A and x∉C iff x∈(A\B) and x∈(A\B) iff x∈ (A\B) n (A\C).

iv) x∈A\(B ∩ C) iff x∈A and x∉B∩C iff x∈A and x∉B or x∉C (if x∈B and x∈C then x∈B∩C thus one of these two must be false) iff x∈A and x∉B or x∈A and x∉C iff x∈(A\B) or x∈(A\B) iff x∈ (A\B) ∪ (A\C).

8 0
3 years ago
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Well to answer question 43, you would have to multiply the radius but 2 so 8*2= 16 and 16* 3.14= 50.24
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