Answer:
3 = 234
Step-by-step explanation:
156 ÷ 2 = 78 so then 156 + 78 = 234
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Answer:
The equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0
Step-by-step explanation:
A line is a polynomial function of the first degree that has the following form:
y = m * x + b
where m is the slope of the line and b is the intercept with the Y axis. The slope is the inclination of the line with respect to the abscissa axis.
In this case you know that the slope of line a is ½.
Parallel lines are two or more lines in a plane that never intersect. Two lines are parallel if they have the same slope and different y-intercepts.
You have:
- y+½x=2 → y= -½x +2
- -2x+y+2=0 → y=2x -2
- 2x-2=y
The line that has the same slope as line a is the last one, so <u><em>the equation of a line that is parallel to line a is y=½x + 2, or what is the same -½x+y-2=0</em></u>
Y-intercept, it can be described as y-intercept form, or another way of writing it in mathematical form is y=mx+b, this can be used to draw polygons, find ortho-centers (very difficult), etc. m is the rise/run, rise over run, or 1/2. The rise and the run is the rate of change. b is the starting point or slope where it starts at, that is most of it!
Answer:
The x-coordinate is \dfrac{\pi}{6}[/tex].
Step-by-step explanation:
We are given a function f(x) as:
Now on differentiating both side with respect to x we get that:
When 
this means that 
Hence, cosine function takes the negative value in second and third quadrant but we have to only find the value in the interval ![[0,\pi]](https://tex.z-dn.net/?f=%5B0%2C%5Cpi%5D)
.
also we know that 
----(1) (which lie in the second quadrant)
so on comparing our equation with equation (1) we obtain:
Hence, the x-coordinates where for 
is 
.
Answer:
Explanation: A midsegment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side, and the length of the midsegment is half the length of the third side.
Hopefully I could Help :)
