A point moves on the x-axis in such a way that its velocity at time t (t>0) is given by v=lnt/t. At which value of t does v attain its maximum? (A) 1
(B) e^(1/2)
(C) e
(D) e^(3/2)
(E) There is no maximum value for v.
Answers
v = ln(t) / t
max of v when dv/dt = 0
dv/dt = (t/t - ln(t))/t²
= [1 - ln(t)]/t² = 0
ln(t) = 1
t = e (option C)
check this is a max and not a min
d²v/dt² = [t²(-1/t) - 2t(1 - ln(t))]/t^4
= [-1 - 2t + 2t ln(t)]/t^4
= [-1 - 2e + 2e ln(e)]/e^4
= [-1 - 2e + 2e]/e^4
= -1/e^4 < 0 so it is a max
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Evaluate b^2 c ^-1 for b = 8 and c = -4. -16 1/16 16 256
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Step-by-step explanation:
- -10w<20
- w<20/-10
- w<-2
hope it helps
stay safe healthy and happy...
Answers
Answer:
150%
Step-by-step explanation:
15 / 10 =
1.5 =
150%
A trinomial has a degree of 2
A constant has a degree of 1
A cubic monomial has a degree of 3
I will mark as the top answers of you help me!
Answers
Answer:
The degree of a binomial is zero. The product of two binomials is not a polynomial. The sum of two polynomials is a polynomial. A monomial containing ^2 has a degree of three
The degree of the polynomial is found by looking at the term with the highest exponent on its variable(s). Examples: 5x2-2x+1 The highest exponent is the 2 so this is a 2nd degree trinomial.
Names of Degrees
Degree Name Example
0 Constant 7
1 Linear x+3
2 Quadratic x2−x+2
3 Cubic x3−x2+5
The degree of a cubic monomial is three. A quadratic polynomial is a trinomial. The degree of a binomial is two.
Step-by-step explanation:
B. 0.75
C. 0.9
D. 0.925
E. 1
Answers
Answers
3x = -30
Divide each side by 3 :
x = -10
Answers
I did it, I got an answer that IS . Here are both of my procedures. If all you want is
the answer, look down below at the second one. But if you could help me out, now
that you know how to do this stuff, please look at my first solution and tell me where
I messed up. I can't find it.
=======================================================
Here's what the problem tells you:
D = 10 log ( ' I ' / 10⁻¹² )
D = 60 . . . . . find ' I ' .
Here we go:
60 = 10 log ( ' I ' / 10⁻¹² )
Divide each side by 10 :
6 = log ( ' I ' / 10⁻¹² )
Raise 10 to the power of each side of the equation:
10⁶ = ' I ' / 10⁻¹²
Multiply each side by 10¹² :
10¹⁸ = ' I ' That's 10^18. It looks bad, because that isn't one of the choices.
Let's try a slightly different procedure:
============================================
After substituting 10⁺¹² for I₀ , we're working with this formula:
D = 10 log ( 'I' / 10⁺¹² )
Let's just look at the log part of that.
The log of a fraction is [ log(numerator) - log(denominator) ]
log of this fraction is [ log( 'I' ) - log(10⁻¹²) ]
But log(10⁻¹²) is just (-12) .
So the log of the fraction is [ log( 'I' ) + 12 ]
And the whole formula is now:
D = 10 [ log( 'I' ) + 12 ]
60 = 10 [ log( 'I' ) + 12 ]
Divide each side by 10 :
6 = log( 'I' ) + 12
Subtract 12 from each side :
-6 = log ( ' I ' )
' I ' = 10⁻⁶
That's choice-'B' .
==================================================
I'm going to leave the first solution up there, in hopes that you, or one
of the many aces, experts, and geniuses that prowl this site constantly,
can weigh in and show me my blunder on the first attempt.