Factor this problem 6x^2-5x-1>0

Mathematics

1 answer:

Greeley [361]3 years ago

3 0

6x^2-5x-1>0

(6x-1)(x+1)

x=1/6 x=-1

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The half-life of a certain radioactive material is 78 hours. An initial amount of the material has a mass of 790 kg.. Write an e

adoni [48]

<span> half life i= 78 hours
amount after 78 hours is 395 kg:
395 = 790e^(k*78)

Dividing by 790 and taking natural log
ln (395/790) = (k*78)
-0.6931 = 78k
-0.00888 = k

lets calculate how much is left after 18 hours:
Amount(18)
= 790e^(-0.00888*18)
Amount = 673.301 kg
hope this helps</span>

4 0

3 years ago

Read 2 more answers

Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently

garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

$\small \sum_{i = 1}^{k} L_i= 2\pi$

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.

we get :

$\small \sum_{i = 1}^{k} C_i.l = 2\pi$

where C(i) is a constant coefficient obviously between 0 and 1.

$\small \sum_{i = 1}^{k} C_i= 2\pi/l$

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]

So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.

We already know the sum so it is easy to compute the average :

$\small L_{Exp} = \frac{2\pi}{k}$