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pychu [463]
3 years ago
8

Combine the like terms to create an equivalent expression. 7k - 3k + 11

Mathematics
1 answer:
Semenov [28]3 years ago
6 0
I believe the answer is 4k + 11
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Two angles are supplementary. The first angle has 20 degreesmore the second angle. How ma y degrees are in each angle?
Vanyuwa [196]

Answer:

100 and 80

Step-by-step explanation:

Let x = be the first angle

x-20 is the second angle

They are supplementary so they add to 180

x+x-20 = 180

Combine like terms

2x-20 =180

2x-20+20 =180+20

2x= 200

Divide by 2

2x/2 = 200/2

x= 100

The first angle is 100 and the second is x-20

x-20 = 100-20=80

The two angles are 100 and 80

3 0
3 years ago
LOGARITHMIC functions have 2 asymptotes<br> True<br> False
JulsSmile [24]

true, both have a domain limited to x values great then 0

8 0
2 years ago
8. Aimi decided to spend her money and save some in the ratio of 3:2. If her total money is RM
Bogdan [553]

Step-by-step explanation:

2/5×7500

2×1500

3000

RM=3000

3 0
3 years ago
Calculate a length of a square ground of perimeter 160m​
wlad13 [49]
Square Perimetre = Side x 4
So a length of a square ground is 160/4 = 40 (m)
5 0
3 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
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