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3 years ago

Find the Least Common Multiple of 15w and 6w^2.

2 answers:

7 0

For this case, we have that by definition, the LCM of two or more natural numbers, is the smallest natural number that is a common multiple of all of them.

Then the LCM of 15 and 6 is the smallest positive integer that divides the numbers 15 and 6 without leaving a residue.

Multiples of each number:

15: 15,30,45 ...

6: 6,12,18,24,30

So, the LCM is $30w ^ 2$

Answer:

$30w ^ 2$

Lina20 [59]3 years ago

3 0

15= 1,3,5,15

6^2= 36

36=1,2,3,4,6....

LCM of 15W and 6W^2=3

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What aspect of the town are most at risk for damage by the fire??

Stolb23 [73]

Answer:

Trees.

Step-by-step explanation:

The houses look sturdy, and not that flammable. The trees and grass are very plentiful, so they would spread easily.

Please clarify if this wasn't what you were asking.

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3 years ago

In the graph shown below, if lines 1^1 are parallel, then the slope of the line 1^2 is?

Galina-37 [17]

Answer:

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7 0

3 years ago

<img src="https://tex.z-dn.net/?f=%24a%2Ba%20r%2Ba%20r%5E%7B2%7D%2B%5Cldots%20%5Cinfty%3D15%24%24a%5E%7B2%7D%2B%28a%20r%29%5E%7B

riadik2000 [5.3K]

Let

$S_n = \displaystyle \sum_{k=0}^n r^k = 1 + r + r^2 + \cdots + r^n$

where we assume |r| < 1. Multiplying on both sides by r gives

$r S_n = \displaystyle \sum_{k=0}^n r^{k+1} = r + r^2 + r^3 + \cdots + r^{n+1}$

and subtracting this from $S_n$ gives

$(1 - r) S_n = 1 - r^{n+1} \implies S_n = \dfrac{1 - r^{n+1}}{1 - r}$

As n → ∞, the exponential term will converge to 0, and the partial sums $S_n$ will converge to

$\displaystyle \lim_{n\to\infty} S_n = \dfrac1{1-r}$

Now, we're given

$a + ar + ar^2 + \cdots = 15 \implies 1 + r + r^2 + \cdots = \dfrac{15}a$

$a^2 + a^2r^2 + a^2r^4 + \cdots = 150 \implies 1 + r^2 + r^4 + \cdots = \dfrac{150}{a^2}$

We must have |r| < 1 since both sums converge, so

$\dfrac{15}a = \dfrac1{1-r}$

$\dfrac{150}{a^2} = \dfrac1{1-r^2}$

Solving for r by substitution, we have

$\dfrac{15}a = \dfrac1{1-r} \implies a = 15(1-r)$

$\dfrac{150}{225(1-r)^2} = \dfrac1{1-r^2}$

Recalling the difference of squares identity, we have

$\dfrac2{3(1-r)^2} = \dfrac1{(1-r)(1+r)}$

We've already confirmed r ≠ 1, so we can simplify this to

$\dfrac2{3(1-r)} = \dfrac1{1+r} \implies \dfrac{1-r}{1+r} = \dfrac23 \implies r = \dfrac15$

It follows that

$\dfrac a{1-r} = \dfrac a{1-\frac15} = 15 \implies a = 12$

and so the sum we want is

$ar^3 + ar^4 + ar^6 + \cdots = 15 - a - ar - ar^2 = \boxed{\dfrac3{25}}$

which doesn't appear to be either of the given answer choices. Are you sure there isn't a typo somewhere?

7 0

2 years ago

Find the area of a circle with radius r = 6 cm

BlackZzzverrR [31]

The formula to find the area of a circle is pi x radius^2.

pi = 3.14 (in this case)
radius = 6

6^2 (6 x 6) = 36

36 x 3.14 (radius^2 x pi) = 113.04

So, the area of a circle with radius 6 cm is 113.04 cm.

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2 years ago

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ipn [44]

Answer:

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3 years ago