Answer:
D
Step-by-step explanation:
7 x 6 = 42 divided by 2 21 + 7 x 5 = 35 divided by 2 = 17.5 + 10 x 7 = 70 = 108.5
Negative reciprocal means
-1/4 times x=-1 solve for x
multiply both sides by -4/1
x=-4
so slope=-4
y=mx+b
m=slope, b=y intercpet
y=-4x+b
input 3 for x and -5 for y
-5=-4(3)+b
-5=-12+b
add 12 to both sides
7=b
the equation in slope intercept form is y=-4x+7
that is my standard form
another standard form is
y+4x=7
Answer:
The easiest way to do this is to reflect Figure I over the Y axis. Then move it up 3 and left 2.
Answer by BlueSky06
The equation described above can also be written as, y = -x² + 100x + 4000To get the number of notebooks that will give them the maximum profit, we derive the equation and equate to zero. dy/dx = -2x + 100 = 0The value of x from the equation is 50. Then, we substitute 50 to the original equation to get the profit. y = -(50^2) + 100(50) + 4000 = 6500Thus, the maximum profit that the company makes is $6,500/day.
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Part A
Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.
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Part B
It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.
dh/dt = (25)/(2pi*h^2)
2pi*h^2*dh = 25*dt
int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides
(2/3)pi*h^3 = 25t + C
2pi*h^3 = 3(25t + C)
h^3 = (3(25t + C))/(2pi)
h^3 = (75t + 3C)/(2pi)
h^3 = (75t + C)/(2pi)
h = [ (75t + C)/(2pi) ]^(1/3)
Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0
0 = [ (75(0) + C)/(2pi) ]^(1/3)
0 = [ (0 + C)/(2pi) ]^(1/3)
0 = [ (C)/(2pi) ]^(1/3)
0^3 = (C)/(2pi)
0 = C/(2pi)
C/(2pi) = 0
C = 0*2pi
C = 0
Therefore the h(t) function is...
h(t) = [ (75t + C)/(2pi) ]^(1/3)
h(t) = [ (75t + 0)/(2pi) ]^(1/3)
h(t) = [ (75t)/(2pi) ]^(1/3)
Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)
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Part C
Your answer is correct.
Below is an alternative way to find the same answer
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Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
75t = 1024pi
t = 1024pi/75
At this time value, the height of the water is 8 feet
Set up the radius r(t) function
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B
Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before
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Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)
