All categories

3 years ago

Find the least common multiple of the following polynomials 9(x+2)(2x-1) and 3(x+2)

Mathematics

1 answer:

ololo11 [35]3 years ago

3 0

A common multiple is a number that is a multiple of two or more numbers. The common multiples of 3 and 4 are 0, 12, 24, .... The least common multiple (LCM) of two numbers is the smallest number (not zero) that is amultiple of both.
so to the given 9(x+2)(2x-1) and 3(x+2)
the LCM would be 9(x+2)(2x-1)

You might be interested in

Greg has 21 new magazines to read. Let M be the number of magazines he would have left to read after reading R of them. Write an

Rina8888 [55]

M = 21 - R is our equation. Plug your value of 12 into the equation for R.

M = 21 - 12.

M = 9.

Greg would have 9 magazines left after reading 12 of them

7 0

3 years ago

G(x) = -0.5x^2 + 4x – 2

-BARSIC- [3]

Answer:

Step-by-step explanation:

What is the question?

g(x) = -0.5x² + 4x - 2 is a down-opening parabola. Do you need to know how to put it in vertex form?

vertex (4,6)

focus (4,5.5)

5 0

2 years ago

Help me pls i have been stuck on this for the past hour

Sedbober [7]

Answer:

-4, -6, -3, -5, -1. The inequality solved for n is n ≥ -6.

Step-by-step explanation:

Substitute all the values in the equation.

n/2 ≥ -3

-10/2 ≥ -3

-5 is not ≥ -3.

n/2 ≥ -3

-7/2 ≥ -3

-3.5 is not ≥ -3.

n/2 ≥ -3

-4/2 ≥ -3

-2 is ≥ -3.

n/2 ≥ -3

-9/2 ≥ -3

-4.5 is not ≥ -3.

n/2 ≥ -3

-6/2 ≥ -3

-3 is ≥ -3.

n/2 ≥ -3

-3/2 ≥ -3

-1.5 is ≥ -3.

n/2 ≥ -3

-8/2 ≥ -3

-4 is not ≥ -3.

n/2 ≥ -3

-5/2 ≥ -3

-2.5 is ≥ -3.

n/2 ≥ -3

-2/2 ≥ -3

-1 is ≥ -3.

To solve the inequality n/2 ≥ -3 for n, do these steps.

n/2 ≥ -3

Multiply by 2.

n ≥ -6.

3 0

3 years ago

Help me please please

Studentka2010 [4]

Answer:

It's going to be B and E because it's in the hundreds place not in the tens or whole

6 0

2 years ago

Consider a system with one component that is subject to failure, and suppose that we have 115 copies of the component. Suppose f

castortr0y [4]

Answer:

Step-by-step explanation:

From the given information:

the mean $(\mu) = 115 \times 20$

= 2300

Standard deviation = $20 \times \sqrt{115}$

Standard deviation (SD) = 214.4761

TO find:

a) $P(x > 3500)= P(Z > \dfrac{3500-\mu}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{3500-2300}{214.4761})$

$P(x > 3500)= P(Z > \dfrac{1200}{214.4761})$

$P(x > 3500)= P(Z >5.595)$

From the Z-table, since 5.595 is > 3.999

$P(x > 3500)=1-0.9999$

P(x > 3500) = 0.0001

Here, the replacement time for the mean $(\mu) = \dfrac{0+0.5}{2}$

= 0.25

Replacement time for the Standard deviation $\sigma = \dfrac{0.5-0}{\sqrt{12}}$

$\sigma = 0.1443$

For 115 component, the mean time = (115 × 20)+(114×0.25)

= 2300 + 28.5

= 2328.5

Standard deviation = $\sqrt{(115\times 20^2) +(114\times (0.1443)^2)}$

= $\sqrt{(115\times 400) +(114\times 0.02082249}$

= $\sqrt{(46000) +2.37376386}$

= $\sqrt{(46000) +(2.37376386)}$

= $\sqrt{46002.374}$

= 214.482

Now; the required probability:

$P(x > 4125) = P(Z > \dfrac{4125- 2328.5}{214.482})$

$P(x > 4125) = P(Z > \dfrac{1796.5}{214.482})$

$P(x > 4125) = P(Z >8.376)$

$P(x > 4125) =1- P(Z$

From the Z-table, since 8.376 is > 3.999

P(x > 4125) = 1 - 0.9999

P(x > 4125) = 0.0001

7 0

2 years ago