The simplified form of (2x+1)(3x−2)/24x 2−4x −8

1 answer:

8 0

$\frac{(2x+1)(3x-2)}{24x^{2}-4x-8} = \frac{(2x+1)(3x+2)}{4(6x^{2}-x-2)} = \frac{(2x+1)(3x+2)}{4(6x^{2}-4x+3x-2)} = \frac{(2x+1)(3x+2)}{4[2x(3x-2)+(3x-2)]} = \\ \\ = \frac{(2x+1)(3x+2)}{4(2x+1)(3x-2)} = \frac{3x+2}{4(3x-2)} = \frac{3x+2}{12x-8}$

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The children from a football club are put into rows in the sports hall. When put into rows of 9 children, there are 2 children l

g100num [7]

We are told that the children from a football club are put into rows in the sports hall. When put into rows of 9 children there are 2 children left over. When put into rows of 12 children there are 2 children left over.

We will find least number of children in football club by finding LCM of 9 and 12.

Multiples of 9 are: 9, 18, 27, 36, 45, 54, 63,...

Multiples of 12 are: 12, 24, 36, 48, 60, 72, 84,...

We can see that least common multiple of 9 and 12 is 36.

We are told that 2 children left over from putting them into 9 and 12 children per row. To find least number we will add 2 to 36.

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How do measurements of time differ for events in a frame of reference that moves at 50% of the speed of light relative to us? At

ZanzabumX [31]

Answer with explanation:

Relation between Speed , Distance and time

Distance =Speed × Time

→It means Speed is inversely Proportional to time.

As distance will remain constant , in that frame of reference

If speed of light in a Medium

$=s=3 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = t seconds or hours or another unit of time.

Now, If speed of light is 50% of the speed of light relative to us

That is Speed of light in another medium

$=w=\frac{50}{100} \times 3 \times 10^8\\\\=1.5 \times 10^8 \frac{\text{meter}}{\text{second}}$

Then, time taken to cross the medium = 2t seconds or hours or another unit of time.

Using Unitary Method

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