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inessss [21]
3 years ago
14

Let the smallest of 4 consecutive odd numbers be 2 n + 1 , where n is an integer. Show, using algebra, that the sum of any 4 con

secutive odd numbers is always a multiple of 8. Give your answer as an expression that makes explicit that it is a multiple of 8.
Mathematics
1 answer:
andrey2020 [161]3 years ago
7 0

Answer:

Expression is 8(n+2)

Step-by-step explanation:

smallest of 4 consecutive odd numbers =2n + 1

consecutive odd integers are found by adding 2 to the any given odd numbers

Thus, 2nd consecutive odd integers  = 2n + 1 + 2 = 2n + 3

3rd consecutive odd integers  = 2n + 3 + 2 = 2n + 5

2nd consecutive odd integers  = 2n + 5 + 2 = 2n + 7

Thus, 4 consecutive odd integers are

2n + 1 ,2n + 3,2n + 5,2n + 7

sum of these numbers are = 2n + 1 +2n + 3 + 2n + 5+2n + 7 = 8n+16

sum of these numbers are = 8(n+2)

Thus, we see that the sum of numbers are 8(n+2)

As, 8 is common for n+2, whatever is value of n, the number will be multiple of 8 .

thus expression is 8(n+2)

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write the standard form of the equation of the circle for which the endpoints of a diameter are (0,0) and (4,-6)
Elenna [48]

Given:

The endpoints of a diameter are (0,0) and (4,-6).

To find:

The equation of the circle.

Solution:

The endpoints of a diameter are (0,0) and (4,-6). So, the length of the diameter is

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

d=\sqrt{(-6-0)^2+(4-0)^2}

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d=\sqrt{52}

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Now, radius is half of the diameter.

r=\dfrac{2\sqrt{13}}{2}

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Center of the circle is the midpoint of the endpoints of a diameter.

Center=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

Center=\left(\dfrac{0+4}{2},\dfrac{0+(-6)}{2}\right)

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Standard form of a circle is

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Therefore, the standard form of the circle is (x-2)^2+(y+3)^2=13.

5 0
2 years ago
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