If the lengths of two sides of a certain triangle are 5 and 10, what is the length of the third side of the triangle?
2 answers:
Answer: The length of the third side is greater than 5 and less than 15 units.
Step-by-step explanation: Given that the lengths of two sides of a certain triangle are 5 and 10 units.
We are to find the length of the third side of the triangle.
Let x represents the length of the third side of the given triangle.
We know that the sum of the lengths of two sides of a triangle is always greater than the length of the third side, so we must have
and
From inequalities (i), (ii) and (iii), we get
Thus, the length of the third side is greater than 5 and less than 15 units.
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Problem 4
<h3>Answer:</h3>------------------
Work Shown:
The left line goes through (-2,0) and (0,4)
The slope of this line is
m = (y2-y1)/(x2-x1)
m = (4-0)/(0-(-2))
m = (4-0)/(0+2)
m = 4/2
m = 2
The y intercept is b = 4
Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.
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The line on the right side goes through (1,-2) and (2,-1)
Slope
m = (y2-y1)/(x2-x1)
m = (-1-(-2))/(2-1)
m = (-1+2)/(2-1)
m = 1/1
m = 1
The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.
Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3
So y = mx+b turns into y = 1x+(-3) or just y = x-3
We combine both parts to end up with
This is only graphed when (note the closed or filled in circle for the endpoint of this portion).
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Problem 5
Answer:
<h3>------------------
Work Shown:
y = |x| .... parent function
y = |x+3| ... shift 3 units to the left
y = (1/2)*|x+3| .... vertically compress by factor of 1/2
f(x) = (1/2)*|x+3|
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Break that down into a piecewise function
when x < -3, then y = -(1/2)(x+3)
when , then y = (1/2)(x+3)
I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when
So that is how we get as the piecewise function.