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3 years ago

The equations y = 5x and y = –1∕5x + 4 represent what types of lines?

Mathematics

1 answer:

docker41 [41]3 years ago

3 0

Answer:

It might be perpendicular lines

Step-by-step explanation:

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Write a power represented with a positive base and a positive exponent whose value is less than the base

Murrr4er [49]

That's always true if either the exponent or the base is less than ' 1 '.

(But not always of they both are.)

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3 years ago

What are the coordinates of the vertex of the graph of

vesna_86 [32]

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3 years ago

What is the area of the red shaded sector ?

scoundrel [369]

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3 years ago

Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured across a line (x + 2)/3 = (2y + 3)/4

dimaraw [331]

Answer:

Distance of the point from its image = 8.56 units

Step-by-step explanation:

Given,

Co-ordinates of point is (-2, 3,-4)

Let's say

$x_1\ =\ -2$

$y_1\ =\ 3$

$z_1\ =\ -4$

Distance is measure across the line

$\dfrac{x+2}{3}\ =\ \dfrac{2y+3}{4}\ =\ \dfrac{3z+4}{5}$

So, we can write

$\dfrac{x-x_1+2}{3}\ =\ \dfrac{2(y-y_1)+3}{4}\ =\ \dfrac{3(z-z_1)+4}{5}\ =\ k$

$=>\ \dfrac{x-(-2)+2}{3}\ =\ \dfrac{2(y-3)+3}{4}\ =\ \dfrac{3(z-(-4))+4}{5}\ =\ k$

$=>\ \dfrac{x+4}{3}\ =\ \dfrac{2y-3}{4}\ =\ \dfrac{3z+16}{5}\ =\ k$

$=>\ x\ =\ 3k-4,\ y\ =\ \dfrac{4k+3}{2},\ z\ =\ \dfrac{5k-16}{3}$

Since, the equation of plane is given by

x+y+z=3

The point which intersect the point will satisfy the equation of plane.

So, we can write

$3k-4+\dfrac{4k+3}{2}+\dfrac{5k-16}{3}\ =\ 3$

$=>6(3k-4)+3(4k+3)+2(5k-16)\ =\ 18$

$=>18k-24+12k+9+10k-32\ =\ 18$

$=>\ k\ =\dfrac{13}{8}$

So,

$x\ =\ 3k-4$

$=\ 3\times \dfrac{13}{8}-4$

$=\ \dfrac{7}{4}$

$y\ =\ \dfrac{4k+3}{2}$

$=\ \dfrac{4\times \dfrac{13}{8}+3}{2}$

$=\ \dfrac{19}{4}$

$z\ =\ \dfrac{5k-16}{3}$

$=\ \dfrac{5\times \dfrac{13}{8}-16}{3}$

$=\ \dfrac{-21}{8}$

Now, the distance of point from the plane is given by,

$d\ =\ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}$

$=\ \sqrt{(-2-\dfrac{7}{4})^2+(3-\dfrac{19}{4})^2+(-4+\dfrac{21}{8})^2}$

$=\ \sqrt{(\dfrac{-15}{4})^2+(\dfrac{-7}{4})^2+(\dfrac{9}{8})^2}$

$=\ \sqrt{\dfrac{225}{16}+\dfrac{49}{16}+\dfrac{81}{64}}$

$=\ \sqrt{\dfrac{1177}{64}}$

$=\ 4.28$

So, the distance of the point from its image can be given by,

D = 2d = 2 x 4.28

= 8.56 unit

So, the distance of a point from it's image is 8.56 units.

4 0

3 years ago

You decide to play roulette 4 times, each time betting on red. What is the distribution of X, the number of times you win?

aalyn [17]

1.
If betting on red, 18/38 = 9/19 chance of winning, so 10/19 chance of losing
Chance of winning 4 times: (9/19)^4 = 6561/130321 =~ 5.034%
Chance of winning 3 times: (9/19)^3 * (10/19) * C(4,1) = 29160/130321 =~ 22.3755%
Chance of winning 2 times: (9/19)^2 * (10/19)^2 * C(4,2) = 48600/130321 =~ 37.2925%
Chance of winning 1 time: (9/19) * (10/19)^3 * C(4,3) = 36000/130321 =~ 27.624%
Chance of winning 0 times: (10/39)^4 = 10000/130321 =~ 76.7336%

3 0

3 years ago