Answer:
Step-by-step explanation:
The square root of 8 to the nearest 100th is 2.83
The 3 is less than 5 so it does not alter the 8 at all.
The answer is 2.8
Answer:
Numbers that aren't: 35 and 43. It can literally be any number.
Perfect cubes: 8,64,125,216. You can find perfect cubes by cubing any number. For example 6^3 (6 to the power of 3) is 216. A perfect cube.
Answer:
59.5
Step-by-step explanation:
1st tree to shadow ratio: 17:10
Plug the numbers in to get x:35
17:10=x:35
17/10=x/35
Multiply both sides by 35
595/10=x
59.5=x
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.
Integer, natural, whole, rational, irrational, natural (15/3=5), integer (-12/6=-2), rational, irrational, rational, rational
Natural-the counting numbers (1,2,3,100...)
Whole-all natural AND 0
Integers-add negative numbers (no fractions)
Rational-add fractions and decimals
Irrational-numbers that can’t be a fraction (square root of 3, pi)
Real-all numbers
