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3 years ago

If n and k are positive integers, is n divisible by 6 ? (1) n = k(k + 1)(k – 1) (2) k – 1 is a multiple of 3.

Mathematics

1 answer:

xz_007 [3.2K]3 years ago

4 0

Answer: yes, it is

Step-by-step explanation:

A number is divisible by 6 if it is divisible by 2 and 3 simultaniously.

n = k(k+1)(k-1)

If k-1 is a multiple of 3, n is divisible by 3, so one of the requirements is ok.

Now, if k-1 is a multiple of 3, it can be even or odd.

if k-1 is even, then it is divisible by 2 and as it is divisible by 3 as well, n is divisible by 6

if k-1 is odd, then k and k+1 is even, hence, divisible by 2.

As n = k(k+1)(k-1), n is also divisible by 6.

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An "honest" coin gets heads 50% of the time and tails the other 50%.

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3 years ago

What is the greatest common factor of 10, 20, and 50?

Viefleur [7K]

Answer:

Step-by-step explanation:

first you need to find the prime factors of the three numbers

ex:

prime factor of 18= $2*3*3$

prime of 24= $2*2*2*3$

there is one 2 and one 3 in common

greatest common multiple 2*3=6

so the greatest common multiple is ten because it is the only one that can divide between all three evenly.

$10=10/10=1\\20=20/10=2\\50=50/10=5$

so 10 is your answer

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3 years ago

Angie has a collection of 145 stuffed animals. She gave 48 to her

Alekssandra [29.7K]

Angie will now have 100 stuffed animals

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3 years ago

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Veronika [31]

The expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Given an integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ .

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

$\int\limits^b_a {f(x)} \, dx$ = $\lim_{n \to \infty}$ ∑f(a+iΔx)Δx ,here Δx=(b-a)/n

$\int\limits^5_1 {x/(2+x^{3}) } \, dx$ =f(x)= $x/2+x^{3}$

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)= $[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}$

$\lim_{n \to \infty}$ ∑f(a+iΔx)Δx=

$\lim_{n \to \infty}$ ∑ $n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n$

=4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$

Hence the expression of integral as a limit of Riemann sums of given integral $\int\limits^5_b {1} \, x/(2+x^{3}) dx$ is 4 $\lim_{n \to \infty}$ ∑ $n(n+4i)/2n^{3}+(n+4i)^{3}$ from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

#SPJ4

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2 years ago

Can someone plz help me on this I’ll appreciate it thank you sooo much

andreyandreev [35.5K]

Answer:

Step-by-step explanation:

Nope<3;p

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3 years ago