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3 years ago

Find the least common multiple of 18 ,12 , and .14

Mathematics

2 answers:

netineya [11]3 years ago

8 0

First, factor all these numbers:

18=2·3·3;
12=2·2·3;
14=2·7.

Then

find all common factors for three numbers - this is factor 2 and write it as first factor of LCM(18,12,14);
find all common factors for two numbers - this is factor 3 and write it as second factor of LCM(18,12,14);
write all remaining factors.

At last, the least common multiple LCM(18,12,14)=2·3·3·2·7=252.

Answer: 252.

sladkih [1.3K]3 years ago

6 0

Hello!
I'm assuming that the values you gave are 18, 12 and 14, but there's a period before the 14, I think that's a mistake but if it's not please let me know, anyhow, the least common multiple of 12, 14 and 18 is 252.

Good luck!

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Given a point translated from A(1,2) to B(4,4). If a point C at (0,0) is translated in the same way, what will be its new endpoi

FinnZ [79.3K]

Answer:

B.(3,2)

Step-by-step explanation:

Point translated from A(1,2) to B(4,4).

This means that for x, we add 3 units(4 - 1 = 3), while for y, we add two units(4 - 2 = 2).

If a point C at (0,0) is translated in the same way, what will be its new endpoints?

0 + 3 = 3

0 + 2 = 2

B.(3,2)

3 0

3 years ago

(21gh^4)(6g^6h^2)(3g^2h^6) find the greatest common factor of the monomials

mezya [45]

The answer is
$3g {h}^{2}$
To find greatest common factor, look at the variables: g, h

Look at their coefficients and exponents:
$21g \: \: \: 6 {g}^{6} \: \: \: 3 {g}^{2} \\ {h}^{4} \: \: \: {h}^{2} \: \: \: {h}^{6}$
Find the two groups' GCF:
$3g \\ {h}^{2}$
Combine:
$3g {h}^{2}$

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3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

lara31 [8.8K]

Answer:

The sample size is 1875.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

Sampling error of 0.03.

This means that $M = 0.03$

99.74% confidence level

So $\alpha = 0.0026$ , z is the value of Z that has a pvalue of $1 - \frac{0.0026}{2} = 0.9987$ , so $Z = 3$ .

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

What is the sample size

The sample size is n. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 3\sqrt{\frac{0.25*0.75}{n}}$

$0.03\sqrt{n} = 3\sqrt{0.25*0.75}$

Simplifying by 0.03 both sides

$\sqrt{n} = 100\sqrt{0.25*0.75}$

$(\sqrt{n})^2 = (100\sqrt{0.25*0.75})^2$

$n = 1875$

The sample size is 1875.

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3 years ago

Help!?!?!?!??!?!?!?!?!??!

Sunny_sXe [5.5K]

24/k
Now, plug in 3 for k
24/3 = 8
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3 years ago

Read 2 more answers

How do you do pythagorean theorem when you need to fine the A

fredd [130]

Given that Pythagorean Theorum is
A^2 + B^2 = C^2, you need to do the opposite.
To find for A^2, you need to:
Subtract B^2 from C^2. This creates the equation of:
C^2 - B^2 = A^2. This works because A^2 + B^2 is C^2, so the opposite must work.
Now if you're looking for A and not A^2, you need to subtract B^2 from C^2 and then square root the difference.
I hope this helps!

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4 years ago