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Sunny_sXe [5.5K]
3 years ago
12

What is the length of MQ?

Mathematics
1 answer:
uranmaximum [27]3 years ago
6 0

Answer:

In general, IBM MQ object names can be up to 48 characters long

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Which of the following shows the graph of y = 4x + 3?
gladu [14]
It would be the graph with the line at the 3 on the y axis and with points going up 4 and over 1 

4 0
3 years ago
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There are currently 441 dairy cows at Dancing Dairy Farm. Due to some considerations, the number of dairy cows is decreasing at
Ymorist [56]

Answer:

the production of milk at the Dancing Dairy Farm is increasing by 2158 gallons/year

Step-by-step explanation:

Given the data in the question;

Let us represent the number of cows at the farm with x and

each cow produces y gallons of milk

Total mil production will be T.

so

T = x × y

now, differentiating with respect to t

dT/dt = d( xy )/dt

dT/dt = xdy/dt + ydx/dt

given that;

x = 441

y = 1157

dx/dt = -13

dy/dt = 39

so we substitute

dT/dt = ( 441 )( 39 ) + ( 1157 )( -13 )

dT/dt = 17199 - 15041

dt/dT = 2158

Therefore, the production of milk at the Dancing Dairy Farm is increasing by 2158 gallons/year

3 0
3 years ago
What is the length of segment DC?
andrew11 [14]

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Step-by-step explanation:

4 0
4 years ago
The​ half-life of a certain tranquilizer in the bloodstream is 47 hours. How long will it take for the drug to decay to 93​% of
GaryK [48]

Answer:

It will take 4.84 hours for the drug to decay to 93​% of the original​ dosage.

Step-by-step explanation:

We are given that the half-life of a certain tranquilizer in the bloodstream is 47 hours.

The given exponential model is: A = A_0 e^{kt}

Now, we know that A becomes half after 47 hours which means that;

A = 0.5 A_0

Using this in the above equation we get;

A = A_0 e^{kt}

0.5 A_0 = A_0 e^{(k\times 47)}  where t = 47 hours

\frac{0.5 A_0}{A_0}  =  e^{(47k)}

0.5 = e^{47k}

Taking log on both sides we get;

ln(0.5) = ln(e^{47k})

ln(0.5) =47k

k = \frac{ln(0.5)}{47}

k = -0.015

Now, the time it will take for the drug to decay to 93​% of the original​ dosage is given by;

0.93 = e^{kt}  where t is the required time

0.93 = e^{(-0.015 \times t)}

Taking log on both sides we get;

ln(0.93) = ln(e^{-0.015t})

ln(0.93) =-0.015t

t = \frac{ln(0.93)}{-0.015}

t = 4.84 hours

Hence, it will take 4.84 hours for the drug to decay to 93​% of the original​ dosage.

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