For how many ordered pairs of positive integers (x,y) is 2x+3y<6
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2
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7
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For how many ordered pairs of positive integers (x,y) is 2x+3y<6
1
2
3
5
7
(0, 0)Quote:
Originally Posted by RK29
(0, 1)
(1, 0)
(1, 1)
(2, 0)
I count 5.
Line l passes through origin and is perpendicular to the line 4x+y=k, where K is constant. If the two lines interesect at ( t, t+1), what is the value of t?
so did I but the answer key is it's 1Quote:
Originally Posted by rdtedm
Answer key is wrong - I have seen many errors in test prep books when I tutor confused students.. Answer has to be five unless it was transcribed wrongQuote:
Originally Posted by RK29
okay thanks
Line l passes through origin and is perpendicular to the line 4x+y=k, where K is constant. If the two lines interesect at ( t, t+1), what is the value of t?
Passes through origin means line has form y= mx, where m is some real number.Quote:
Originally Posted by RK29
Rearranging the perpendicular line, we get y = -4x + k, so this line has a slope of -4, so the inverse reciprocal slope is 1/4.
Which means, Line 1 has an equation of y = (1/4)x
Line 1: y = (1/4)x
Line 2: y = -4x + k
Since we know they intersect at (t, t+1), this must be a solution to both equations. Since Line 2 would have infinite solutions (since k is unknown), we choose Line 1 to solve for t.
t+1 = (1/4)t
1 = (-3/4)t
(-4/3) = t
So, t = -4/3
Quote:
Originally Posted by RK29
The answer key is correct.Quote:
Originally Posted by rdtedm
There is only one pair of positive integers that works: (1,1).
@ rdtedm, you should that 0 is not a positive integer.
Hello, RK29;669895!
Quote:
. .\,1 \qquad (b)\;2 \qquad (c)\;3 \qquad(d)\;5 \qquad (e) 7)
For positive integers, there is one pair:
Ah, Plato beat me to it . . .Code:|
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