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BigorU [14]
2 years ago
15

The sum of two numbers is 24. Their difference is 15. What are the two numbers?

Mathematics
1 answer:
ipn [44]2 years ago
5 0
X+y = 24x-y = 15 Thus, x=15+y Substitute for x in first equation becomes 15 +y + y = 24. 2y=9; therefore y=4.5, and since x = 24-y, x= 19.5 Check answer: 4.5 + 19.5 = 24; 19.5-4.5 = 15

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\bf f(x)=y=2x+sin(x)
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inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
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\textit{now, the "y" in the inverse, is really just g(x)}
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\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
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\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
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g'(2)=\cfrac{1}{2+cos[g(2)]}

now, if we just knew what g(2)  is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that  means if we use that value in f(x),   f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus    2 = 2x+sin(x)

\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
5 0
2 years ago
If AK= 14, EK=17, BK= 7 , What is the length of DK? 12.7 8.5 7.0 3.5
Elden [556K]
To get the value of DK we use proportionality:
AK/EK=BK/KD
thus plugging the values we get:
14/17=7/KD
getting the reciprocal of getting both sides we have:

17/14=KD/7
thus
KD=17/14×7
KD=8.5
thus 
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