Mathematics, 10.09.2020 03:01 deborah0

Hey! Can you help out a bit?

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Answer from: jose477

$k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ \perp\ l\iff m_1m_2=-1\\====================================\\k:y=-x-2;\ l:y=mx+b\\\\k\ \perp\ l\iff-1\cdot m=-1\Rightarrow m=1\\\\l:y=1x+b\to y=x+b\\\\(4;\ 1)\to substitute\ x=4\ and\ y=1\ to\ y=x+b:\\\\4+b=1\\b=1-4\\b=-3\\\\y=x-3$

Answer from: GreenHerbz206

Hey ! I'll try.

You haven't said what the question is, and it doesn't show in the picture.

I'll assume that you're asked to find the equation of the line perpendicular to
the given line and passing through the point (4, 1). Hope that's it.

First, you need to remember that perpendicular lines have negative reciprocal slopes.

The given line has slope of -1, so the line perpendicular to it has slope of +1/1 = 1 .

So the equation of the new line is going to be [ y = 1x + intercept ].

You know that the new line goes through the point where x=4 and y=1.
Stick these into the part of the equation that you already know:

y = 1x + intercept

1 = 1(4) + intercept.

Can you find the intercept from here ?
Maybe I'd better just finish it off.

1 = 4 + intercept.

Subtract 4 from each side:

intercept = -3

So the equation of the line perpendicular to [ y = -x - 2 ] going through (4, 1) is

y = x - 3

Answer from: Quest

the range of a function is the range of y-values for which it is defined, or exists. given that we are only looking at the y-values, we can already discount the first two answers as they deal with the domain of the function (range of x-values).

now, what we need to do is find the smallest and highest y-values for which the function is defined. if you'd like, you could take a ruler and start at the very bottom of the graph, find the lowest point, then move the ruler higher and higher until you reach the highest point. doing so either in your head or with an actual ruler, we can see that the lowest y-value is -5 and the highest y-value is -1.

the question now remains as to whether they are included in the range (fourth answer) or excluded (third answer). the key here is to recognise whether they are marked with an open or a closed circle. since both (-2, -5) and (2, -1) are marked with a closed circle, they are points that exist on the graph and are thus included in the range (note that if they were marked with an open circle like (4, -2) then that would signify that there is a break in the graph at that point).

thus, the fourth answer, (y | -5 ≤ y ≤ -1), is correct.

also note that usually it is good practice to scan your eye inbetween the highest and lowest point for any breaks in the graph - we do have one here at (4, -2), however -2 is still included in the range because there is another point that is included in the graph with the same y-value, ie. point (1, -2).

i hope that but if you require any further explanation don't hesitate to comment below.

Me if you can and if you finished this lesson on edg- comment you.

Answer from: Quest

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer

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