2228.
]]>Bob
]]>Excellent! It's a problem I do with most of my maths classes at some time, a part of my topic of "The Nature of Proof". You can do it at any age and with very little prep.
I think it's important as it shows that just getting a recognisable sequence is not enough to be certain of the 'formula'.
Other readers: Don't look up the answer; try it for yourself; there's useful learning lurking here.
Bob
]]>- OscarMathius
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Teachers open the door, but you have to enter it yourself - Chinese proverb
Based on the energy conservation, we can write (since the exercise is about an ideal situation):
m*v^2/2 = (m+M)*g*h
therefore:
h = m*v^2/(m+M)/g/2
(Va) = 12600 liters (was 16,800 liters)
AB = 360 miles (was 420 miles)
(Vt) = 4200 liters (as before)
(Ct) = 10 liters/mile (as before)
For example, 15x + 12y = 500 has no solutions because the GCD (or HCF) of 15 and 12 is 3, which doesn't divide 500.
And 17x + 29y = 1000 must have a solution in the integers because the GCD (or HCF) of 17 and 29 is 1, which divides 1000.
Also, once you've found a solution (x, y), then all the other solutions have the form
for an arbitrary integer k.
You may like to solve for example:
107*A + 109*B = 10,000
You can extend this trick to find integer solutions to much more complicated equations (e.g. quadratics, cubics) but you'd need to do a bit more work. For example, you can take modulo 4 of both sides of the equation
to show that this doesn't have any integer solutions.
]]>[(S2*T1) + R]^2 = D^2 + (S1*T)^2 , //T1 = time from W to X//
(S2*T1)^2 + 2*S2*T1*R + R^2 = D^2 + (S1*T)^2
(S2*T1)^2 + 2*S2*T1*R + R^2 - D^2 - (S1*T)^2 = 0
If T2 = time from X to Y, T2=R/S3
T = T1 + T2
T1 = T – T2
T1 = T – R/S3
By replacing T1:
S2^2*(T – R/S3)^2 + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0
S2^2*(T^2 – 2*R/S3*T + R^2/S3^2) + 2*S2*(T – R/S3)*R + R^2 - D^2 - (S1*T)^2 = 0
S2^2*T^2 – 2* S2^2*R/S3*T + S2^2*R^2/S3^2 + 2*S2*R*T – 2*S2*R^2/S3 + R^2 - D^2 - S1^2*T^2 = 0
(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + S2^2*R^2/S3^2 – 2*S2*R^2/S3 + R^2 - D^2 = 0
(S2^2 - S1^2)*T^2 – 2* S2*R*(S2/S3 - 1)*T + R^2*(S2/S3 – 1)^2 - D^2 = 0
This is a quadratic equation whose general parameters (a), (b) and (c) are:
a = S2^2 - S1^2
b= –2* S2*R*(S2/S3 - 1)
c= R^2*(S2/S3 – 1)^2 - D^2
A numeric example:
D = 24.11 km
S1 = 30 km/h
S2 = 45 km/h
S3 = 150 km/h
R = 20 km
The mission time is T = 15 min.
Kerim
]]>Anyway, I guess you meant by your last line that the condition to be satisfied by the parameters A, B, C and D for f(x) doesn’t have local high and low limits is:
B^2 - 3*A*C ≤ 0
And the parameter (D) may have any value.
Thank you.
]]>Bob
]]>1. Solve this equation: 3x-2+-6=4(x²+2x)2.
what is 2. at end? a square?
2. Solve this equation: ln(4x)=(x-cos(x²))(3x-2)3.
what is 3. at end? (i dont think u can solve this w/ algebra; have to use numeric approx)
3. create an linear system of equations where x=2,y=0 and z=2.5 4.
pick numbers for x,y,z. use what they gave you to get values.
e.g. 3x+2y+4z=3*2+2*0+4*2.5 = 6+0+10 = 16 so eqn is 3x+2y+4z=16
now do two more to get system
Let a³x²+b²x+c=0. What values are a,b,c if a < b and c = 4a
put 4a in for c: a^3x^2+b^2x+4a=0
do quad fiormula for x in terms of a&b
do cases. e.g. if x=0 then what? (a=0 and b>0 so infinite sol'n set)
try other cases
(https://www.wolframalpha.com/input?i=a%5E3+x%5E2+%2B+b%5E2+x+%2B+4a+%3D+0%2C+a+%3C+b)
]]>And here's the link to the recording I used: Gerhard Gschossmann - 'Summertime' (George Gershwin) guitar solo fingerstyle
That version is played at the song's correct tempo, with good feeling and expression...which makes my Geogebra soundtrack sound rather chipmunkeyish.
Here's a vocal clip of the song: GEORGE & IRA GERSHWIN 'Summertime' HAROLYN BLACKWELL ('Porgy & Bess'). Very moving!!
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