Answer:
1/7 (option d) of the sensors on the satellite have been upgraded
Step-by-step explanation:
Each unit contains the same number of non-upgraded sensors
number of non-upgraded sensors for each module (nus)
total number of upgraded sensors on the satellite (tus)
satellite is composed of 30 modular units
total number of non-upgraded sensors on the satellite (tnus):
tnus=30*nus
total number of sensors on the satellite (ts):
ts=tnus+tus = 30*nus + tus (I)
The number of non-upgraded sensors on one unit is 1/5 the total number of upgraded sensors on the entire satellite
nus=(1/5)*tus
tus = 5 * nus (II)
Fraction of the sensors on the satellite have been upgraded (FU):
FU = tus/ts
Using I and II
FU= (5* nus)/(30*nus + tus)
FU = (5* nus)/(30*nus + 5 * nus)
FU = (5* nus)/(35*nus)
FU = 1/7
1/7 (option d) of the sensors on the satellite have been upgraded
Answer:
a = 6
x^2 + 6x is equal to x(x+6)
b=2
Denominator and numerator of the first term are multiplied by x.
c=6
Second term is multiplied by (x-6)/(x-6)
d=2
Now that they have the same denominator, the two terms are combined. 2 is the coefficient of the first term
e=6
In the same way as d is carried over from b, e is carried over from c.
f = 6
2x - x + 6 = x + 6
g = 1
We factor out the (x+6) from the numerator and denominator
Answer:
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Step-by-step explanation:
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Answer:
t ≅ 5.09 min
Step-by-step explanation:
we have that if in 4000 L/sol there is 132 kg salt and the pumping speed is 12L/s, we must find how much of salt is pumping per second and then find the amount of salt remaining
12L/s*132kg salt/4000L = 0.396 Kg salt/s, this means that 0.396 kg per second comes out
, It should be found that the amount of salt must be drained so that only 11 kg of salt remain
132kg salt - 11 kg salt = 121 kg salt, so
121Kg salt*s/0.396Kg salt ≅ 305.55 s ⇒ 305.55s*min/60s ≅ 5.09 min
