Lines <em>a</em> and <em>b</em> are parallel, so lines <em>p</em>, <em>q</em>, and <em>t</em> are considered to be transversals. To solve this, you make use of the fact that alternate interior angles are equal, as are alternate exterior angles, as are corresponding angles. Of course any linear pair of angles is supplementary.
∠1 = 90° — corresponding angle to the right angle above it
∠2 = 68° — the sum of 22° and angles 1 and 2 is 180°
∠3 = 112° — supplementary to angle 2 (and the sum of 22° and 90°, opposite interior angles of the triangle)
∠4 = 112° — equal to angle 3
∠5 = 68° — equal to angle 2; supplementary to angle 4
∠6 = 56° — base angle of isosceles triangle with 68° at the apex; the complement of half that apex angle
∠7 = 124° — supplementary to the other base angle, which is equal to angle 6; also the sum of angles 5 and 6
∠8 = 124° — alternate interior angle with angle 7, hence its equal.
Answer:
On an average, the shop 1 has a lighter variety of laptops.
False
If you buy a laptop of 10 pounds, then it's more likely that you bought the laptop from shop 1.
False
Shop 2 has more laptops with a weight less than the average weight than does Shop 1.
True
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
Slope formula: y2-y1/x2-x1
= 12-5/3-2
= 7/1
= 7
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Best Regards,
Wolfyy :)
Answer:
In geometry, an obtuse scalene triangle can be defined as a triangle whose one of the angles measures greater than 90 degrees but less than 180 degrees and the other two angles are less than 90 degrees.
