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3 years ago

Proves that the sum of 5 consecutive natural numbers is divisible by 5

Mathematics

2 answers:

erastova [34]3 years ago

6 0

Answer:

n = 5m + 10 = 5(m + 2), since m + 2 is an integer, then the n is divisible by 5.

Step-by-step explanation:

borishaifa [10]3 years ago

5 0

Answer:

Natural numbers are simply positive integers. For now, refer to the middle number (the third number), as x. Thus, the numbers are: x - 2, x - 1, x, x + 1, and x + 2. Simply add all these numbers to get 5x, a number that is divisible by 5. As such, you can conclude that any 5 consecutive numbers are divisible by 5, and the quotient would be the third, or middle number.

Hope it helps :) and let me know if you want me to elaborate.

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I NEED AN ANSWER ASAP, THANKS

Naddik [55]

V = 4/3πr^3V = 4/3 (π) (6.5)^3
V = 366.1666π (6 repeat)
answer is 366.16 π (3rd choice)

7 0

3 years ago

A 3 cm x 3 cm rectangle sits inside a circle with radius of 4 cm.

goblinko [34]

Answer:

41.24cm²

Step-by-step explanation:

See attachment for the figure

First is to find area of rectangle

Given:

Length of rectangle 'l'= 3cm

breadth of rectangle 'b'= 3cm

Area of rectangle= l x b= 3 x 3=> 9cm²

Next is to find the area of the circle

Given:

radius 'r'= 4cm

Area of circle= πr² => π4² =>50.24cm²

Required:

area of the shaded region=?

area of the shaded region= area of the circle - area of the rectangle

area of the shaded region= 50.24 -9 =>41.24cm²

Therefore, area of the shaded region is 41.24cm²

8 0

3 years ago

Select ONE of these problems to solve. STAR the problem and then write the expression first. After you write the expression, sol

Sveta_85 [38]

Answer:

Step-by-step explanation:

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James - 3x sneekers

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7 0

2 years ago

Read 2 more answers

Determine whether the series is absolutely convergent 1-(1*3/3! (1*3*5/5!-（3*5*7）

DENIUS [597]

It's not clear what your series is, so I'm going to take a wild guess on what it is you mean:

$1-\dfrac{1\times3}{3!}+\dfrac{1\times3\times5}{5!}-\dfrac{1\times3\times5\times7}{7!}+\cdots$
$=\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(2n-1)!}\prod_{k=1}^n(2k-1)$

For the sum to be absolute convergent, the sum of the absolute value of the summand must converge, so you are really examining the convergence of

$\displaystyle\sum_{n=1}^\infty\frac1{(2n-1)!}\prod_{k=1}^n(2k-1)$

This is easily checked with the ratio test:

$\displaystyle\lim_{n\to\infty}\left|\frac{\displaystyle\dfrac1{(2(n+1)-1)!}\prod_{k=1}^{n+1}(2k-1)}{\displaystyle\dfrac1{(2n-1)!}\prod_{k=1}^n(2k-1)}\right|=\lim_{n\to\infty}\left|\frac{\dfrac{1\times3\times5\times\cdots\times(2n-1)\times(2n+1)}{(2n+1)!}}{\dfrac{1\times3\times5\times\cdots\times(2n-1)}{(2n-1)!}}\right|$
$\displaystyle=\lim_{n\to\infty}\left|\frac{\dfrac{2n+1}{(2n+1)(2n)}}{\dfrac11}\right|=\lim_{n\to\infty}\dfrac1{2n}=0$

Since $\sum|a_n|$ converges by the ratio test, the series $\sum a_n$ converges absolutely.

4 0

3 years ago

The fuel tank of a certain airplane can hold up to gallons of fuel. Let be the total weight of the airplane (in pounds). Let be

zloy xaker [14]

A reasonable function is W = 7F + 5000

Where W is the weight of the plane and F is the number of gallons of fuel.

The domain is the set of possible values for F. That is from 0 (empty tank) to 400 (full tank).

So the domain is F = [0,400], or what is equivalente 0 ≤ F ≤ 400.

The range is the set of values of W.

The minimum value of W is when F = 0 => W = 7(0) + 5000 = 5,000.

The maximum value of W is when F = 400 => W = 7(400) + 5000 = 7,800

So the range is W = [5,000 ; 7,800], pr 5000 ≤ W ≤ 7,800.

3 0

2 years ago

Proves that the sum of 5 consecutive natural numbers is divisible by 5 ​

Proves that the sum of 5 consecutive natural numbers is divisible by 5