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Zepler [3.9K]
3 years ago
5

I need help on a math problem

Mathematics
1 answer:
timofeeve [1]3 years ago
5 0
What is the problem?

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Would someone help me understand this type of math
Svetlanka [38]
I never understand that
6 0
3 years ago
Brainliest for whoever gets this right
meriva

Answer

14 degrees

Step-by-step explanation:

7 0
2 years ago
Could someone please explain I keep getting the wrong answer
Naddika [18.5K]
Of course! so basically for each value of x, for example -6, plug it into the equation. so for -6 do -2/3(-6)+7. hope this helps!
3 0
2 years ago
What is a necessary step for constructing perpendicular lines through a point off the line?
Nostrana [21]

Answer:

Find another point on the perpendicular line.

Step-by-step explanation:

Given an original line "m", and a point off the line "Q", in order to construct a second line "p", meant to be perpendicular to "m" through the point "Q", fundamentally, the only truly necessary step to construct a perpendicular line through is to find another point on the yet-to-be-found perpendicular line.

Most often, this is accomplished by exploiting the fact that "p" is the set of all points that are equidistant from any pair of points that are symmetric about "p".

Since the symmetry must be about "p", and we don't even know where "p" is, one often finds two points on "m" that are equidistant from "Q".

This can be accomplished by adjusting a compass to a fixed radius (larger than the distance from "Q" to "m"), and making an arc that intersects "m" in two places.  Those two places will be equidistant from "Q", and are simultaneously on line "m".  Thus, these two points, "A" & "B" are symmetric about "p".

Since "A" & "B" are symmetric about "p", they are equidistant from "p", and are on "m".  One could try to find the point of intersection between "p" and "m" through construction, but this is unnecessary.  We need only find a second point (besides "Q") that is equidistant from "A" & "B", which will necessarily be a point on "p", to form the line perpendicular to "m".

To do this, fix the compass with any radius, and from "A" make a large arc generally in the direction of "B", and make the same radius arc from "B" in the direction of "A" such that the two arcs intersect at some point that isn't "Q".  This point of intersection we can call point "T", and the line QT is line "p", the line perpendicular to the original line, necessarily containing "Q".

8 0
2 years ago
Which equation has x=2/3 as a solution and why?
eduard

There are an infinite number of different equations that have
x=2/3  as the solution, or as one of several solutions.

You didn't show us your list of choices, so we can't help you
pick the right one.

Just take each choice, one at a time.  Write ' 2/3 ' in place of 'x'
in each equation, and simplify everything.  You'll wind up with
one equation that's true, and the others will say something that's
nonsense. 

The one that's true is the one that has  x=2/3  as its solution.

6 0
3 years ago
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