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Vinvika [58]
3 years ago
5

Lee was 2 under par on the first hole, 3 under par on the second and third hole, 2 over par on the fourth, 5 over par on the fif

th, and he made par exactly on the sixth and seventh holes. He was one under par on the eighth hole and 1 over par on the ninth. How many strokes above/below par was Lee after the first nine holes? Express your answer as an integer.
Mathematics
1 answer:
Anika [276]3 years ago
3 0
1 under par overall
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for that two lines be parallel the condition is  that  two slope be equal m1 = m2

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m = -5-0/6-(-7) , m = -5/13

for the seconds points R (-1,-1) and S( 0,-4)

m = -4-(-1)/ 0-(-1) = -3/1 = -3

the lines are not parallel, m1 ≠ m2

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A. Use composition to prove whether or not the functions are inverses of each other. B. Express the domain of the compositions u
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Given: f(x) = \frac{1}{x-2}

           g(x) = \frac{2x+1}{x}

A.)Consider

f(g(x))= f(\frac{2x+1}{x} )

f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}

f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}

f(\frac{2x+1}{x} )=\frac{x}{1}

f(\frac{2x+1}{x} )=1

Also,

g(f(x))= g(\frac{1}{x-2} )

g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}

g(\frac{1}{x-2} )= \frac{x }{1}

g(\frac{1}{x-2} )= x


Since, f(g(x))=g(f(x))=x

Therefore, both functions are inverses of each other.


B.

For the Composition function f(g(x)) = f(\frac{2x+1}{x} )=x

Since, the function f(g(x)) is not defined for x=0.

Therefore, the domain is (-\infty,0)\cup(0,\infty)


For the Composition function g(f(x)) =g(\frac{1}{x-2} )=x

Since, the function g(f(x)) is not defined for x=2.

Therefore, the domain is (-\infty,2)\cup(2,\infty)



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4 years ago
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