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3 years ago

What is the least common multiple of 16, 6, and 12?

2 answers:

5 0

6: 6,12,18,24,30,36,42,<u>48</u>,54,60
12: 12,24,36,<u>48</u>
16: 16,32,<u>48</u>

Answer: 48

First you write down all the multiples of 6,12 and 16 then you find the first number that appears in each of the lists of numbers or the one that is the least common multiple.

Aleonysh [2.5K]3 years ago

5 0

16|2
8|2
4|2
2|2
1

6|2
3|3
1

12|2
6|2
3|3
1

$LCM(6,12,16)=2^4\cdot3=16\cdot3=48$

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4 years ago

Write a rule for the nth term of the arithmetic sequence. Then graph the first six terms of the sequence. a6=-12 a12=-36

denis-greek [22]

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We need to find the first term. We can use the formula

an = a1 + d(n - 1)

to solve for a1. We already know the 12th term, a12, common difference, d, and nth sequence, n.

a12 = -36

d = -4

n = 12

-36 = a1 - 4(12 - 1)

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The first term is 8. Therefore, your formula is

an = 8 - 4(n - 1)

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Then use this formula to graph.

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an is the dependent variable.

Your graph will be a line.

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Step-by-step explanation:

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3 years ago

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

$(x^2-(-4))+(y^2-(-4))=4$

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

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3 years ago