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3 years ago

Calculate the distance from each given point to the given line.

2 answers:

7 0

Given: 2x-y-3=0.
find equation for the line perpendicular to the given line that goes through the given point:
(2;-1)koord of direction vector (i`m not know how it is called at you, because i'm from russia)
=> (x-0)/2=(y-4)/-1 (canonical equation)
=>x+2y-8=0(general equation)

further:
{x+2y-8=0
{2x-y-3 =0 => y=13/5 x=14/5

(14/5; 13/5) - koord point on line
|dist|=sqrt( (14/5-0)^2 + (13/5-4)^2 ) = sqtr(7.72) = 2.78

Удачи!

.

mojhsa [17]3 years ago

3 0

The line y = 2x -3 has a gradient of 2.

so the line perpendicular to it has a gradient of -1/2 and will have the general formula of y = -1/2 x + c.

to find c, use the coordinates (0,4)

4 = 0 + c

so c = 4

equation is y = -1/2 x + 4

If you need the distance of (0,4) from the line you will need to put both lines on a graph to find the intersection (and possibly use simultaneous equations for more accurate answer) and then use pythagoras to find lengths.

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3 years ago

Which statement is true about the factorization of 30x2 + 40xy + 51y2? The polynomial can be rewritten after factoring as 10(3x2

iragen [17]

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The greatest common factor of the terms is 1.

Step-by-step explanation:

The terms have no variables in common, and the coefficients have no factors in common. The greatest common factor of the terms is 1.

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3 years ago

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3 years ago

Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$