❣ R̾e̾f̾e̾r̾ t̾h̾e̾ a̾t̾t̾a̾c̾h̾m̾e̾n̾t̾.... ❣
☺.. h̾o̾p̾e̾ i̾t̾ h̾e̾l̾p̾s̾ y̾o̾u̾.. ☺
y = 9ln(x)
<span>y' = 9x^-1 =9/x</span>
y'' = -9x^-2 =-9/x^2
curvature k = |y''| / (1 + (y')^2)^(3/2)
<span>= |-9/x^2| / (1 + (9/x)^2)^(3/2)
= (9/x^2) / (1 + 81/x^2)^(3/2)
= (9/x^2) / [(1/x^3) (x^2 + 81)^(3/2)]
= 9x(x^2 + 81)^(-3/2).
To maximize the curvature, </span>
we find where k' = 0. <span>
k' = 9 * (x^2 + 81)^(-3/2) + 9x * -3x(x^2 + 81)^(-5/2)
...= 9(x^2 + 81)^(-5/2) [(x^2 + 81) - 3x^2]
...= 9(81 - 2x^2)/(x^2 + 81)^(5/2)
Setting k' = 0 yields x = ±9/√2.
Since k' < 0 for x < -9/√2 and k' > 0 for x >
-9/√2 (and less than 9/√2),
we have a minimum at x = -9/√2.
Since k' > 0 for x < 9/√2 (and greater than 9/√2) and
k' < 0 for x > 9/√2,
we have a maximum at x = 9/√2. </span>
x=9/√2=6.36
<span>y=9 ln(x)=9ln(6.36)=16.66</span>
the
answer is
(x,y)=(6.36,16.66)
<h3>
Answer: D) 3/150</h3>
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Explanation:
With the use of a calculator, we see that,
- 11/19 = 0.57894736842106...., the decimals eventually repeat; but unfortunately my calculator ran out of room to show the repeating portion
- 4/7 = 0.5714285714285714..., the block "571428" repeats forever
- 1/3 = 0.333333.... the 3s go on forever
- 3/150 = 0.02
So 3/150 converts to the terminating decimal 0.02
The word "terminate" means "stop". In the other decimal values, the decimal digits go on forever repeating the patterns mentioned.
----------------------
A non-calculator approach will have us simplify 3/150 into 1/50 after dividing both parts by the GCF 3. Then notice how 50 has the prime factorization of 2*5*5. The fact that the denominator 50 can be factored in terms of only 2's and 5's is enough evidence to conclude that the fraction converts to a terminating decimal.
If the denominator factors into some other primes, other than 2s and 5s, then we don't have a terminating decimal. So that's why 11/19, 4/7 and 1/3 convert to non-terminating decimals.
Let x = the other rational number.
x(66/7) = (48/5)
Solve for x to find your answer.
