Answer:
well it's says for a its 35° and B is 100° to find x it would be 180-(35+110)=x
Step-by-step explanation:
180-(35+110)=180-145=35. so x=35°
Answer: m²-9mn+2n²
Step-by-step explanation: make use of BODMAS strategy. Let the unknown be x.
-7mn+2m²+3n²-x= m²+2mn+n². Since we are concerned about getting the unknown,let's try to collect the like terms.
-7mn+2m²+3n²-(m²+2mn+n²)= x
-7mn+2m²+3n²-m²-2mn-n²= x
-7mn-2mn+2m²-m²+3n²-n²=x
-9mn+m²+2mn= x
X= m²-9mn+2n²
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)
Question:
gbuhhubhhbuubbbubuubhbhhhbbubhbhbhubuhhbhbu
Answer:
yes
Answer:
b. about 91.7 cm and 44.6 cm
Step-by-step explanation:
The lengths of the diagonals can be found using the Law of Cosines.
Consider the triangle(s) formed by a diagonal. The two given sides will form the other two sides of the triangle, and the corner angles of the parallelogram will be the measure of the angle between those sides (opposite the diagonal).
For diagonal "d" and sides "a" and "b" and corner angle D, we have ...
d² = a² +b² -2ab·cos(D)
The measure of angle D will either be the given 132°, or the supplement of that, 48°. We can use the fact that the cosines of an angle and its supplement are opposites. This means the diagonal measures will be ...
d² = 60² +40² -2·60·40·cos(D) ≈ 5200 ±4800(0.66913)
d² ≈ {1988.2, 8411.8}
d ≈ {44.6, 91.7} . . . . centimeters
The diagonals are about 91.7 cm and 44.6 cm.
