The answer is Q.
Suppose you have the points with coordinates (p, q) and (r, s) then, provided p is different from r, the slope of the line is (q - s)/(p - r) = m, say. Then, if (x, y) is any point on the line, (x - s)/(y - r) = m That, after simplification, is the linear equation of the line. This will be a lot simpler when you have numerical values for p, q, r and s rather than work algebraically throughout. If p is not different from r, then the equation is x = p (or r), a vertical line.
stuck between a rock and a hard place
Yes. If one matrix is p*q and another is r*s then they can be multiplied if and only if q = r and, in that case, the result is a p*s matrix.
Two ratios, p/q and r/s (q and s non-zero) are equal if p/q - r/s = 0.
C. P. D. r a i a c n p k o p e s y d a u r s (sponsered by shreck 2 productions)
P=s r t , so, s= P/(st)
p/q * r/s = (p*r)/(q*s)
19999555
A rational number is a number of the form p/q where p and q are integers and q > 0.If p/q and r/s are two rational numbers thenp/q + r/s = (p*s + q*r) / (q*r)andp/q - r/s = (p*s - q*r) / (q*r)The answers may need simplification.
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It is 3*(q + p)/(r + s)
If a is rational then there exist integers p and q such that a = p/q where q>0. Similarly, b = r/s for some integers r and s (s>0) Then a*b = p/q * r/s = (p*r)/(q*s) Now, since p, q r and s are integers, p*r and q*s are integers. Also, q and s > 0 means that q*s > 0 Thus a*b can be expressed as x/y where p and r are integers implies that x = p*r is an integer q and s are positive integers implies that y = q*s is a positive integer. That is, a*b is rational.
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Prove: [ P -> Q AND R -> S AND (P OR R) ] -> (Q OR S) -> NOT, --- 1. P -> Q ___ hypothesis 2. R -> S ___ hypothesis 3. P OR R ___ hypothesis 4. ~P OR Q ___ implication from hyp 1. 5. ~R OR S ___ implication from hyp 2 6. ~P OR Q OR S ___ addition to 4. 7. ~R OR Q OR S ___ addition to 5. 8. Let T == (Q OR S) ___ substitution 9. (~P OR T) AND (~R OR T) ___ Conjunction 6,7 10. T OR (~P AND ~R) ___ Distribution from 9 11. T OR ~(P OR R) ___ De Morgan's theorem 12. Let M == (P OR R) ___ substitution 13. (T OR ~M) AND M ___ conjunction 11, hyp 3 From there, you can use distribution to get (T AND M) OR (~M AND M). The contradiction goes away leaving you with T AND M, which can simplify to T.
yea and its spelled sprouse not sprose s-p-r-o-u-s-e not s-p-r-o-s