All categories

3 years ago

Select all the situations where the equation of a line can be found.

2 answers:

3 0

The following are the situations wherein the equation of the line can be found:

A.) Two points on the line are known
<span>D.) A point on the line and the slope are known.
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.

Talja [164]3 years ago

3 0

D. You can use the point-slope formula, A. You can find the slope (y1-y2/x1-x2) and then use the point-slope formula, and E- same explanation as A.

You might be interested in

Between which two consecutive whole numbers does 5 lie

MrRissso [65]

Your answer is 4 and 6

4 0

3 years ago

A bag contains some white and black balls . The probability of picking two white balls one after other without replacement from

FromTheMoon [43]

Answer:

2/35

Step-by-step explanation:

Let w and b be the numbers of white and black balls in the bag respectively.

So, the total numbers of the balls in the bag is

$n=w+b\;\cdots(i)$

As the bag can hold maximum 15 balls only, so

$n\leq15 \;\cdots(ii)$

Probability of picking two white balls one after other without replacement

=Probability of the first ball to be white and the probability of second ball to be white

=(Probability of picking first white balls) x( Probability of picking 2nd white ball)

Here, the probability of picking the first white ball $=\frac{w}{n}$

After picking the first ball, the remaining

white ball in the bag $= w-1$

and the remaining total balls in the bag $=n-1$

So, the probability of picking the second white ball $=\frac{w-1}{n-1}$

Given that, the probability of picking two white balls one after other without replacement is 14/33.

$\Rightarrow \frac{w}{n} \times \frac{w-1}{n-1}=\frac{14}{33}$

$\Rightarrow \frac{w(w-1)}{n(n-1)} =\frac{14}{33}$

Here, $w$ and $n$ are counting numbers (integers) and 14 and 33 are co-primes.

Let, $\alpha$ be the common factor of the numbers $w(w-1)$ (numerator) and $n(n-1)$ (denominator), so

$\frac{w(w-1)}{n(n-1)} =\frac{14\alpha}{33\alpha}$

$\Rightarrow w(w-1)=14\alpha\cdots(iii)$

And $n(n-1)=33\alpha.$

As from eq. (ii), $n\leq 15$ , so, the possible value of $\alpha$ for which multiplication od two consecutive positive integers (n and n-1) is $33\alpha$ is 4.