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snow_tiger [21]
4 years ago
14

This one is very hard thank you

Mathematics
1 answer:
makvit [3.9K]4 years ago
3 0

check the picture below.


if we run a vertical line test on that graph, like you see there, namely run vertical lines through it, we can see that one vertical line only hits the graph once, not twice or more.


on the vertical line test, if the graph is hit only once by each of the vertical lines, then the graph is a function.

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Prove that 17^16+17^14 is divisible by 5
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We'll use the exponent rule that a^(b+c) = a^b*a^c

to break up 17^16 into 17^(14+2) = 17^14*17^2

That allows us to factor like so

17^16+17^14

17^14*17^2+17^14*1

17^14(17^2 + 1)

Computing 17^14 is a lot more difficult compared to computing 17^2 + 1, assuming you only could use paper and pencil.

Note that 17^2 = 17*17 = 289, so 17^2 + 1 = 289+1 = 290

Because 290 ends in a 0, this means that 5 is a factor

290 = 5*58

This makes 5 to be a factor of the original expression.

----------------

So overall,

17^16+17^14 = 17^14(17^2 + 1) = 17^14*290 = 17^14*5*58 = 5*17^14*58

The stuff "17^14*58" doesn't matter. All we care about is that factor of 5.

Since 5 is a factor, this means that 17^16+17^14 is a multiple of 5, or 17^16+17^14 is divisible by 5.

Phrased another way, 17^16+17^14 all divided over 5 leads to some integer. In this case, the integer would be the result of 17^14*58.

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