Need Help With Math Problem!
06-23-2003, 09:50 AM
#1
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Okay,
I have one math problem left on my quiz that I can't figure out.
The sum of three consecutive odd integers is the same as the smallest of these integers. Find the smallest of these integers and find the greatest of these integers.
__________Smallest ___________Greatest
I have one math problem left on my quiz that I can't figure out.
The sum of three consecutive odd integers is the same as the smallest of these integers. Find the smallest of these integers and find the greatest of these integers.
__________Smallest ___________Greatest
it doesn't make sense to me, because how can you add numbers to a smaller number, and get the smaller number for your answer?
06-23-2003, 10:14 AM
#3
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Originally Posted by TyresmokinRx7' date='Jun 23 2003, 11:04 AM
What the hell is an integers. 
06-23-2003, 10:27 AM
#5
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Originally Posted by rfreeman27' date='Jun 23 2003, 11:16 AM
negative
I think they may have screwed up writing the problem or something...unless someone has an answer that I can't seem to find
06-23-2003, 10:47 AM
#6
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easy
-3 -1 1
-3 + -1 = -4
-4 + 1 = -3
-3 -1 1
-3 + -1 = -4
-4 + 1 = -3
The sum of three consecutive odd integers is the same as the smallest of these integers. Find the smallest of these integers and find the greatest of these integers.
-3 Smallest 1 Greatest
-3 Smallest 1 Greatest
06-23-2003, 11:00 AM
#7
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AN interger is non negative because it lies in the set of real numbers. If you go into negative numbers with intergers you are now entering the set of imaginary numbers. Those problems are brain teasers and the way to solve it by trial and error using odd and even numbers. I just failed my math test for Differential Equations.
SOlve this problem:
x^2 means x squared okay..
y' means first derivative.. '' means second and so on
e= eulers equation = 2.7
ln= natural log
cos t^2 + sin t^2 =1 ( identity)
cot = cotangent
sec= secant
IIGht
Find a Particular Solution for
y'' + 4y= 3 csc t ( csc is the cosecant so its 1/sin (t) )
SO the general solution is yc(t) = C1 cos 2t + C2 sin2t
The general solution is off this form because the discriminat b2-4ac ( part of the quadratic equation of the form -b +- (b2-4ac)^(1/2) / 2(a) ). is less than 0. so then the general equation is y= e ^ lambda (t) ( c1 cos u(t) + c2 sin u(t) ) and the discrimant therfore gives us lambda = 0 so e raised to the zero is 1 and then we get C1 cos 2t + C2 sin2t.
Find the coefficiants c1 and c2 byusing variation of parameters where the first condition that needs to be satisfied is
1) v1'y1 + v2'y2= 0
2) v1'y1' + v2'y2' = csc (t) by using trigonometric substituition. Engineers of NP.com gimme this answer!
The answer is y= 3 sint + 3/2 ln | csc t - cot t|sin2t + c1 cos2t + c2 sin2t
SOlve this problem:
x^2 means x squared okay..
y' means first derivative.. '' means second and so on
e= eulers equation = 2.7
ln= natural log
cos t^2 + sin t^2 =1 ( identity)
cot = cotangent
sec= secant
IIGht
Find a Particular Solution for
y'' + 4y= 3 csc t ( csc is the cosecant so its 1/sin (t) )
SO the general solution is yc(t) = C1 cos 2t + C2 sin2t
The general solution is off this form because the discriminat b2-4ac ( part of the quadratic equation of the form -b +- (b2-4ac)^(1/2) / 2(a) ). is less than 0. so then the general equation is y= e ^ lambda (t) ( c1 cos u(t) + c2 sin u(t) ) and the discrimant therfore gives us lambda = 0 so e raised to the zero is 1 and then we get C1 cos 2t + C2 sin2t.
Find the coefficiants c1 and c2 byusing variation of parameters where the first condition that needs to be satisfied is
1) v1'y1 + v2'y2= 0
2) v1'y1' + v2'y2' = csc (t) by using trigonometric substituition. Engineers of NP.com gimme this answer!
The answer is y= 3 sint + 3/2 ln | csc t - cot t|sin2t + c1 cos2t + c2 sin2t
