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3 years ago

14

Prove that if n is a positive integer, then n is even if and only if 7n 4 is even.

1 answer:

deff fn [24]3 years ago

3 0

<span>We prove first the direct implication. Assume n is even. Then n = 2k for some integer k. Then 7n + 4 = 14k + 4 = 2(7k + 2), which is even. For the converse, which is â€śif 7n + 4 is even, then n is evenâ€ť, we use a proof by contrapositive. The contrapositive is: â€śif n is not even (that is, odd), then 7n + 4 is not even (that is, odd) If n is odd, then n = 2k + 1, for some integer k. Then 7n + 4 = 14k + 11 = 2(7k + 5) + 1, which is odd.</span>

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Answer:

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borishaifa [10]

Answer:

495 combinations of 4 students can be selected.

Step-by-step explanation:

The order of the students in the sample is not important. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

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4 from a set of 12. So

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3 years ago

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choli [55]

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3 years ago

The area of a circle is 161 in. What is the radius?

kirill [66]

Answer:

Approximately 7.16 in

Step-by-step explanation:

Hi there!

Area of a circle equation: $A=\pi r^2$ where r is the radius

Plug in the given area 161 in.²

$161=\pi r^2$

Divide both sides by π to isolate r²

$\frac{161}{\pi} =\frac{\pi r^2}{\pi} \\\frac{161}{\pi} =r^2$

Take the square root of both sides to isolate r

$=r^2\\\sqrt{\frac{161}{\pi}} =r\\7.16=r$

Therefore, the radius of the circle is approximately 7.16 in.

I hope this helps!

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2 years ago

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Answer:

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Step-by-step explanation:

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3 years ago