a scale model of a school building is 5 inches tall. If the building is 100 feet tall, find the scale of the model
2 answers:
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Answers:
- Incorrect
- Correct
- Correct
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Explanation:
When applying any kind of reflections, the parallel sides will stay parallel. Check out the diagram below for an example of this.
So PQ stays parallel to RS. Also, QR stays parallel to PS.
The statement "PQ is parallel to PS" is incorrect because the two segments intersect at point P. This letter "P" is found in "PQ" and "PS" to show the common point of intersection. Parallel lines never intersect.
Answer:
x = 4
Step-by-step explanation:
For an equation of this sort, my first step is to rewrite it so the solutions are the x-intercepts of the graph:
(x +5)^(3/2) -(x +1)^3 = 0
The graph is attached. The one real solution is x=4.
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The solution method for something like this is necessarily ad hoc. Here, it appears that progress can be made by raising both sides of the equation to the 2/3 power:
((x +5)^(3/2))^(2/3) = ((x -1)^3)^(2/3)
x +5 = (x -1)^2
Now, you have an ordinary quadratic that can be solved using any of the usual methods.
x^2 -2x +1 = x +5 . . . . swap sides, expand the square
x^2 -3x -4 = 0 . . . . . . put in standard form
(x -4)(x +1) = 0 . . . . . . factor
The solutions to this are ...
x = 4, x = -1 . . . . . . . values of x that make the factors zero
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To see if either of these solutions is extraneous, we can check them:
(4 +5)^(3/2) = (4 -1)^3 . . . . . substitute 4 for x
9^(3/2) = 3^3 . . . . . . . . . true
(-1 +5)^(3/2) = (-1 -1)^3 . . . . . substitute -1 for x
4^(3/2) = (-2)^3
8 = -8 . . . . . . . . . . . false; x = -1 is extraneous
The solution is x = 4.
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<em>Additional comment</em>
You can eliminate the x=-1 "solution" by considering the domains of the left- and right-side expressions. The 3/2 power is equivalent to the square root of the cube. That will generally only be defined for non-negative values (x ≥ -5). The cube on the right will only be positive for x > 1, so the practical domain of this equation is x ≥ 1.
