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Serhud [2]
4 years ago
11

What are the domain restrictions of a + 1 / a^2-5a-36?

Mathematics
1 answer:
velikii [3]4 years ago
3 0
All real numbers except -4 and 9.    

The domain of an expression is the range of all values for which the expression is defined. Looking at the expression, it will legal everywhere except when a^2 - 5a - 36 is equal to 0. Since that's a quadratic equation, we can use the quadratic formula to find where its value is 0, giving -4 and 9.    

Therefore the domain of the expression is, All real numbers except -4 and 9.
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Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b
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since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0



4 0
3 years ago
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