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2 years ago

Find the LCM of 6! and (4!)^2

Mathematics

2 answers:

KIM [24]2 years ago

5 0

Answer:

2880

Step-by-step explanation:

!=factorial. Factorial means you are multiplying by a number by the numbers less than it.

so that means 6*5*4*3*2*1=720.

And the factorial of 4 is 4*3*2=24.

24^2=576

So now we have to find the LCM of 576 and 720.

I will use the prime factorization method so the prime factorization of 576 is:

2*2*2*2*2*2*3*3

Prime factorization of 720:

2*2*2*2*3*3*5

So we first we gotta find the GCF, which is 2*2*2*2*3*3 because that is something that they both share in their prime factorizations. So the answer to the GCF is 144.

So now, the numbers that we did not use which is a 2 and another 2 and a 5. So 2*2*5=20

<h3><u>Finally, we have to multiply 20 by the GCF, which is 144, and the answer is 2880. </u></h3>

sergiy2304 [10]2 years ago

4 0

Answer:

it's 12

4, 8, <u><em>12</em></u>

6, <u><em>12</em></u>

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3 years ago

A quantity, y, varies directly as x When y=10, x=6.<br> Find xwhen y= 14.

grigory [225]

The value of x when y = 14 is 8.4

<h3 />

<h3>How to solve variation ?</h3>

y, varies directly as x. Therefore,

y ∝ x

Therefore,

y = kx

where

k =constant of proportionality

Therefore,

10 = 6k

k = 10 / 6

k = 5 / 3

Hence, when y = 14

y = kx

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x = 42 / 5

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learn more on variation here: brainly.com/question/20707320

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7 0

2 years ago

A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len

aniked [119]

Answer:

a. V = (20-x) $x^{2}$ $in^{3}$

b . 1185.185 $in^{3}$

Step-by-step explanation:

Given that:

The height: 20 - x (in )
Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h * $x^{2}$ $in^{3}$

<=> V = (20-x) $x^{2}$ $in^{3}$

b. What is the maximum possible volume of the box?

To maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3 $x^{2}$ + 40x

Let dV / Dx = 0, we have:

-3 $x^{2}$ + 40x = 0

<=> x = 40/3

=>the height h = 20/3

So the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 $in^{3}$

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3 years ago