Statement: 1) Triangle ABC
2) 2x 3)x=52
Reason: I’m not sure but probably addition
Hello, again!!
So, the range is the outcome when you plug in any x. In other words, the y.
Because this is absolute value, the x will turn positive. However, the number will then turn negative.
Therefore, y will ALWAYS be negative.
The answer is C.
Hope this helps!! Let me know if you have ANY questions.
Answer:
You will need 20 sides to complete the loop.
Step-by-step explanation:
The question isn't quite clear given how small the corner is, but I assume that we are looking to complete the circle if the pentagon and square are repeated in a loop
We can also see - assuming that those are proper equal-sided polygons, that PQ is the same length as PV
With that in mind, We can solve this by noting that the angle of a corner in a square is 90 degrees, and in a pentagon it's 108 degrees.
108 - 90 is equal to 18. This means that PQ is at eighteen degrees to YP. Also, QM, (which will be equivalent to the next VP is eighteen degrees to PQ.
This means that each polygon is rotated 18 degrees relative to it's neighbour.
With all that we can say that the total polygons we need to form a circle is 360/18 = 20, So you will need 20 polygons, or ten squares and ten pentagons to complete the loop.
There are infinitely many lines that have the point (1,-3).
A line can be expressed as:
y=mx+b, where m=slope and b=y-intercept..
Our only restriction is that it passes through (1,-3) so
-3=1m+b
So as long as the sum of the slope and the y-intercept is equal to -3, that is one of the infinite number of lines that passes through (1, -3)
So we could also say b=-3-m then our infinite lines are:
y=mx-3-m, now any real value of m creates a specific line that passes through the point. ie the first few are
y=x-4, y=2x-5, y=3x-6 or even y=x√2-3-√2
