The law of cosines for [tex]\triangle R S T[/tex] can be set up as [tex]$5^2=7^2+3^2-2(7)(3) \cos (S)$[/tex]. What could be true about [tex]\triangle R S T$[/tex]?
Law of cosines: [tex]$a^2=b^2+c^2-2 b c \cos (A)$[/tex]
A. [tex]$r=5[/tex] and [tex]$t=7$[/tex]
B. [tex]$r=3[/tex] and [tex]$t=3$[/tex]
C. [tex]$s=7[/tex] and [tex]$t=5$[/tex]
D. [tex]$s=5[/tex] and [tex]$t=3$[/tex]
Answer (3)
xy=520 & y=x+6 xy = x*(x+6) = x^2 + 6x x^2+6x-520 = 0
=> x = {-26, 20}
When x = -26, y = -20 x = 20, y = 26
it was the answer the above comment ;
The two numbers are 20 and 26, with 26 being 6 more than 20. Their product is 520. Thus, the solution satisfies the conditions of the problem.
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