y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
Answer:
Suzanne is incorrect.
Step-by-step explanation:
To figure out if the two equations are equivalent, we can simplify each one of them.
Simplifying the first equation, we get:

Simplifying the second equation, we get:

Now that both equations are simplified, we can see that the equations are not equivalent. Suzanne is incorrect.
Answer:
Just find the line with a rise of 1 and a run of 4.
Step-by-step explanation:
To find a line first find a actual point that connects and just go up 1 and go to the right of 4. It has to go right becuase since it’s a positive slope it’d have to go right for the run. FInd a point go up 1 and right 4 and if the results gets you landed on a another point, you have suceded to find the graph with A s lope of 1/4
You take $22,000.00 + $625.00= 22,625.00 (which is the sales tax)
Take $22,625.00 x .06=$1,357.50
Then take $22,625.00 + $<span>1,357.50 +$40.00 = $24.022.50</span>
Answer:
Both fireworks will explode after 1 seconds after firework b launches.
Step-by-step explanation:
Given:
Speed of fire work A= 300 ft/s
Speed of Firework B=240 ft/s
Time before which fire work b is launched =0.25s
To Find:
How many seconds after firework b launches will both fireworks explode=?
Solution:
Let t be the time(seconds) after which both the fireworks explode.
By the time the firework a has been launched, Firework B has been launch 0.25 s, So we can treat them as two separate equation
Firework A= 330(t)
Firework B=240(t)+240(0.25)
Since we need to know the same time after which they explode, we can equate both the equations
330(t) = 240(t)+240(0.25)
300(t)= 240(t)+60
300(t)-240(t)= 60
60(t)=60

t=1