Answer:
i think the answer might be 35 or 15
Step-by-step explanation:
When you reflect a point across the x-axis, the x-coordinate remains the same, but the y-coordinate is transformed into its opposite (its sign is changed). If you forget the rules for reflections when graphing, simply fold your paper along the x-axis (the line of reflection) to see where the new figure will be located.p explanation:
Answer:
A) rational number
Step-by-step explanation:
56 is a rational number because it can be expressed as the quotient of two integers like 56 ÷ 1.
3y = -2x + 2
y = x + 4 (Substitute the y)
3(x + 4) = -2x + 2 (Distribute)
3x + 12 = -2x + 2 (Add 3 to both sides)
+3x +3x
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12 = 1x + 2 (Subtract 2 from both sides)
-2 - 2
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10 = 1x (Divide by 1 on both sides)
10/1 = 1x/1
x = 10
This is your x-coordinate
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.
