Two parallel lines are crossed by a transversal. If m6 = 123.5°, then m1 is 56.5°. 67.5°. 123.5°. 136.5°.
2 answers:
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Answer:
d) one solution; (4, 1)
Step-by-step explanation:
It often works well to follow problem directions. A graph is attached, showing the one solution to be (4, 1).
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You know there will be one solution because the lines have different slopes. Each is in the form ...
y = mx + b
where m is the slope and b is the y-intercept.
The first line has slope -1 and y-intercept +5; the second line has slope 1 and y-intercept -3. The slope is the number of units of "rise" for each unit of "run", so it can be convenient to graph these by starting at the y-intercept and plotting points with those rise and run from the point you know.
<h3>Answer:</h3>
(x, y) ≈ (1.49021612010, 1.22074408461)
<h3>Explanation:</h3>This is best solved graphically or by some other machine method. The approximate solution (x=1.49, y=1.221) can be iterated by any of several approaches to refine the values to the ones given above. The values above were obtained using Newton's method iteration.
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Setting the y-values equal and squaring both sides of the equation gives ...
... √x = x² -1
... x = (x² -1)² = x⁴ -2x² +1 . . . . . square both sides
... x⁴ -2x² -x +1 = 0 . . . . . polynomial equation in standard form.
By Descarte's rule of signs, we know there are two positive real roots to this equation. From the graph, we know the other two roots are complex. The second positive real root is extraneous, corresponding to the negative branch of the square root function.