All categories

4 years ago

Least common multiple of 11 and 18

Mathematics

1 answer:

yaroslaw [1]4 years ago

8 0

Hello there!

Explanations:

↓↓↓↓↓↓↓↓↓↓↓↓↓↓

First you had to find the prime factorization of 11.

$1*11=11$

Factors: 1, 11

Then you had to find the prime factorization of 18.

$3*3*2=18$

Factors: 1, 2, 3, 6, 9, 18

You can also multiply by the each factor the greater number of times and it occurs in step one and step 2.

$11*3*3*2=198$

The least common multiple of 11 and 18 is "198".

Hope this helps!

Thank you for posting your question at here on Brainly.

Have a great day!

-Charlie

You might be interested in

What is the sum of the series? 4 ∑k=1 (2k^2−4)

larisa86 [58]

The sum of the series $\sum_{k=1}^{4}\left(2 k^{2}-4\right)$ is 44.

Step-by-step explanation:

The given series is $\sum_{k=1}^{4}\left(2 k^{2}-4\right)=44$

To find the sum of the series, we need to substitute the values for k in the series.

$\sum_{k=1}^{4}\left(2 k^{2}-4\right)=\left[2(1)^{2}-4\right]+\left[2(2)^{2}-4\right]+\left[2(3)^{2}-4\right]+\left[2(4)^{2}-4\right]$

Now, simplifying the square terms, we get,

$[2(1)-4]+[2(4)-4]+[2(9)-4]+[2(16)-4]$

Multiplying the terms,

$[2-4]+[8-4]+[18-4]+[32-4]$

Subtracting the values within the bracket term, we get,

$-2+4+14+28$

Now, adding all the terms, we get the sum of the series,

$\sum_{k=1}^{4}\left(2 k^{2}-4\right)=44$

Thus, the sum of the series is $\sum_{k=1}^{4}\left(2 k^{2}-4\right)=44$

8 0

3 years ago

Slope of the line with a figure (picture included)

Crank

Answer:

the rise is 1 and i think the run is negative 2 so the slop should be 1/-2

Step-by-step explanation:

6 0

3 years ago

Which of these is a binomial? O A. x + 2y2 - 7 O B. ab O C. 5xy O D. 2x3 - 7y3

Lemur [1.5K]

Answer:

Step-by-step explanation:

2 terms

8 0

3 years ago

Blaise’s school is selling tickets to a fall musical. On the first day of ticket sales they sold 8 adult tickets and 3 student t

dezoksy [38]

Your two systems will be

8A + 3S = 79$

your 2nd will be

10A + 5S = 105$

4 0

3 years ago

If (x)=-2377 +3 and V(x)= x, what is the range of (LOV)(x)?

kipiarov [429]

$L(x)=-2377+3\Longrightarrow L(x)=-2374 \\V(x)=x$

Both of the functions are linear. $L(x)$ is constant.

However we are required to determine the range of $(L\circ V)(x)$

$(L\circ V)(x)=L(V(x))=L(x)=-2374$

So the only possible solution that such notation gives is -2374 so therefore the answer in terms of range is $x\in\{-2374\}$

Hope this helps.

r3t40

3 0

4 years ago