Let u be speed while Lisa drove 240 miles and t be the time in completing 240 miles.
We have been given that Lisa increased her speed 20 mph than initial speed over next 540 miles. So her speed on next 540 miles will be 
mph.
Further more we are given that she took 3 hours more to complete the next 540 miles of the trip than she took in completing initial 240 miles. So she took 
hours in completing next 540 miles.
Since we know 
Now we will substitute our given values in our equation and find out the unknowns. After substitution we will get two equations and two unknowns.
Now we will substitute equation 1 in equation 2
Upon solving this equation, we get
Therefore, total time for Lisa to complete the entire trip is 6+6+3 = 15 hours.
And we know that total distance covered is 240+540 = 780 miles
Hence, average speed
Answer:
The graph is attached.
Step-by-step explanation:
Let be "x" the number of metamorphic samples and "y" the number of sedimentary samples.
You know that you want to to have at most 25 samples, then:
And you want to have at least 3 times as many sedimentary samples as
metamorphic samples, then:
Therefore, the system of inequalities that model the situation is:
The y-intercept of the line 
is:
And the x-intercept is:
Knowing this, you can graph the line
The y-intercept of the line 
is:
And the x-intercept is:
Knowing this, you can graph the line.
Observe the graph attached, where the solution of the system of inequalities is the intersection of the regions.
Answer:
Step-by-step explanation:
We move all terms containing a to the left, all other terms to the right
5a=22
a=22/5
a=4+2/5
Answer:
The answer would be $40.
Step-by-step explanation:
You take the first 25% off and add it to the other 25% off, which would equal 50% off. After that you take 50% off for $80, which would be $40 because half of $80 is $40.
Answer:
1,012 ; 1,001
Step-by-step explanation:
Given the population function :
P(t) = 1000+(12e^−0.34t)
where t is the number of years from the time the species was added to the lake.
Initial population size of the species ; that is the population at year 0
At time, t = 0
P(0) = 1000+(12e^−0.34(0))
= 1000 + (12e^0)
= 1000 + 12(1)
= 1012
Initial population = 1,012
Population after 9 years :
P(9) = 1000+(12e^−0.34(9))
= 1000 + (12e^-3. 06)
= 1000 + (12×0.0468876)
= 1000 + ( 0.5626523)
= 1000.5626
= 1001 ( nearest whole number)
