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Tresset [83]
3 years ago
11

Help me please I really need help please Example ( 3 ) (4) (5)

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
7 0
3)
7168
divisible by 2 because the last digit is divisible by 2
NOT divisible by 3 because the sum of its digit (22) is not a multiple of 3
divisible by 4 because the last 2 digits (68) are divisible by 4
NOT divisible by 5 because the last digit (8) is neither 0 nor 5
NOT divisible by 6 (to be divisible y 6, it should be divisible by 2 and 3)
divisible by82 because the last 3 digits (168) ae divisible by 8
NOT divisible by 9 because the sum of its digit (22) is not a multiple of 9
NOT divisible by 10 because the last digit (8) is not 0

Having written the rules, you can easily apply them to solve the last 2 exercices. 
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Find a polynomial function of degree 3 such that f(10)=17 and the zeros of f are 0, 5 and 8
fomenos

Step-by-step explanation:

Since f(0) = f(5) = f(8) = 0, we have f(x) = Ax(x - 5)(x - 8), where A is a real constant.

We know that f(10) = 17.

=> A(10)(10 - 5)(10 - 8) = 17

=> A(10)(5)(2) = 17

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