Question

Please anyone who's willing to help me with Algebra 2 comment.

Answer 1

hi

Answer 2

which q's r u having difficulty with?

Answer 3

Sorry I'm just responding can you still help me?

Answer 4

:[ please

Answer 5

i can try to help

Answer 6

heya, sorry i went for lunch, bak now

Answer 7

Hi

Answer 8

You guys can help me?

Answer 9

Now it is your turn to plan the trajectories required to launch a spacecraft through a specific route in space. The launch area is identified on the map below


Launch Area:___(1, 2)___
Point A:___(0, 3)___
Point B:___(-3, 0)___
Point C:___(-1, -4)___
Graph the cordinates of the specifics points in space your spacecraft will travel to.
You must show your work on each question below.

Determine the equation of the line, in standard form, that will get your spacecraft from the Launch Area to Point A.

Determine the equation of the line, in point-slope form, that will get your spacecraft from Point A to Point B.

Determine the equation of the line, in slope-intercept form, that will get your spacecraft from Point B to Point C.

Convert the equations you arrived at in question 2 and into slope-intercept form. Make sure to include all of your work.

Reflect back on this scenario and each equation you created. Would any restrictions apply to the domain and range of those equations? Explain your reasoning using complete sentences.


HEEEELP

Answer 10

so do you know how to find the equation of a straight line, if u have 2 points?
note: the equation of a striaght line (in slope-intercept form) is \[\large y = mx + b\]
where: m = gradient of the line
b = where the line crosses the y-axis

Answer 11

so fo this first line, from launch area to point A,
\[\large y = mx + b\]
we can see from the picture that the line crosses the y axis at y = 3
its always at x=0, so if we have the point (0,3), we know straight away that b = 3
so our equation now looks like this:
\[\large y = mx + 3\]
now we just need to find what m is in this case
m = gradient (so slope of the line)
you can see from the pic that the slope is downhill (going from left to right)


and it's changing by -1 y value, for every +1 x value it goes across

Answer 12

so the equation to work out m is always
\[\large m = \frac {y_2 - y_1}{x_2 - x_1}\]

so moving from left to right, your first point is:
\[\large A = (0,3) ~~\text {which is} ~(x_1, y_1)\]
then our 2nd point (moving from left to right) is:
\[\large Launch~Point = (1,2) ~~\text {which is} ~(x_2, y_2)\]
so if \[\large m = \frac {y_2 - y_1}{x_2 - x_1}~ \text {and}~ (x_1, y_1) = (0,3)~\text {and}~ (x_2, y_2) = (1,2) \]
then
\[\large m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {2 - 3}{1 - 0} = \frac {-1}1 = -1 \]
so m = -1 and b = 3
so the final equation for the line is:
\[\large y = mx + b\]\[\large \color{red} {y = -1x + 3}\]

Answer 13

now u try for:
Determine the equation of the line, in point-slope form, that will get your spacecraft from Point A to Point B. @Jajajaeel , i'll be back online tomorrow to help and see how u went

Answer 14

ok im here