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2 years ago

What is LCM of 18,24,36,45

1 answer:

5 0

To obtain the least common multiple (l.c.m), we will do it by canonical decomposition or simultaneous decomposition.

This method consists of extracting the common and uncommon prime factors, then

$\large\displaystyle\text{$\begin{gathered}\sf \left.\begin{matrix} \blue{18 \ \ \ 24 \ \ \ 36 \ \ \ 45 }\\ \ 9 \ \ \ 12 \ \ \ 18 \ \ \ 45\\ \ 9 \ \ \ \ \ 6 \ \ \ \ \ 9 \ \ \ 45\\ \ 9 \ \ \ \ \ 3 \ \ \ \ \ 9 \ \ \ 45\\ \ 3 \ \ \ \ \ 1 \ \ \ \ \ 3 \ \ \ 15\\ \ 1 \ \ \ \ \ 1 \ \ \ \ \ 1 \ \ \ \ 5\\ \ 1 \ \ \ \ \ 1 \ \ \ \ \ 1 \ \ \ \ 1 \end{matrix}\right|\begin{matrix} 2\\ 2\\ 2\\ 3\\ 3\\ 5\\ \: \end{matrix} \end{gathered}$}$

$\sf{L.c.m.(18,24,36,45)=2\times2\times2\times3\times3\times5 }$

$\sf{L.c.m.(18,24,36,45)=2^{3} \times3^{2} \times5 }$

$\boxed{\boxed{\sf{L.c.m.(18,24,36,45)=360 }}}$

Therefore, the least common multiple of 18, 24, 36 and 45 is 360.

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I need help with rational and irrational numbers here is the question.

4vir4ik [10]

Answer:

The correct answer choices are:

1) It is a real number

2) It is an irrational number

Step-by-step explanation:

So we have the number:

$\sqrt2$

First, let's define each of the terms:

Real numbers are all of the irrational and rational numbers. If a number is irrational or rational, then they are a real number. Any number that does not include the imaginary number i is real (you'll learn this eventually. It's actually really interesting!).

Irrational numbers are the numbers that do not terminate nor repeat. Examples include π or √3. Note that numbers like 1/3 which is .3333 is <em>not </em>irrational because they are repeating. π, on the other hand, does not repeat and never ends.

Rational numbers are the numbers that terminate or repeat. They also include the integers, whole numbers, and natural numbers.

So, the √2. First, it's real because it doesn't deal with imaginaries.

Next, use a calculator and calculate √2. We can see that it is about 1.41423562... This number does not repeat. And it does not terminate (end). In other words, √2 is a irrational numbers.

Since it is irrational, it cannot be rational. Since it's not rational, it cannot be an integer, whole number, or a natural number.

Therefore, the correct answer choices are:

1) It is a real number

2) It is an irrational number

3 0

3 years ago

Read 2 more answers

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$