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whats the max of this function f(x)=(x^3/3)-4x^2+12x+1

The maximum is where the first derivative is 0 and the second derivative is less than 0.  

\(f(x)=(x^3/3)-4x^2+12x+1\\ f'(x)=x^2-8x+12\\ f''(x)=2x-8\\~\\ x^2-8x+12=0\\ (x-2)(x-6)=0\\ x=+2\;\; or \;\;x=+6\\ y''(2)=4-8<0 \quad \text{Max at x=2}\\ y''(6)=12-8>0 \quad \text{Min at x=6}\\ When\;\;x=2\\ f(2)=8/3-4*4+12*2+1\\ f(2)=8/3-16+24+1\\ f(2)=8/3+9\\ f(2)=11\frac{2}{3}\\ \text{So there is one (local) maximum point at }(2,11\frac{2}{3}) \)

Of course the limit of this function as x tends to infinity is infinity so there are an infinite number of function values that are  higher than this maximum. :)

5,976

Press the  ' 2nd' key before your function key....

45e0

For really big numbers like this, it is easiest to use Scientific notation:

2.5 x 10^9    x  35 x 10^6  =  87.5 x 10^15  = 8.75 x 10^16

35n + 12 = 82

Subtract 12 from both sides:

82-12=70

Now divide 35n/70 

N= 2

Nope..   It's a 'condiment'

If Randall parents do not charge him any interest, then all 4 scenarios are not possible, because:

A - $850 - (5 x $94)  =$380 in debt.

B - $850 - (6 x $94)  =$286 in debt.

C - $850 - (7 x $94)  =$192 in debt.

D - $850 - (8 x $94)  =$98   in debt.

Simple multibplication 0f 26 rows x 18seats/row = 468 seats     (see how the 'row' cancels out??)

9/12 x 1500 x  3/100% =

1125 x .03 = 33.75

Right. So therefore, its 2,350,000,000,000

=  a^k * a^k / 2 a^k  =     a^k / 2

Subtract 4 x from both sides of the equation to get

3x+2 = -10    now, subtract 2 from each side

3x = -12       Divide both sides by 3

x = -4

Well it'd be 2.35*10¹² RIght?

Professor Freeman likes to to think that he has more time to finish his work than he really has. So, he has set the four clocks in his study to be an average of exactly 2 hours slow. If 2 of the professor's clocks read 6:55, a third reads 6:42 and the fourth reads 6:28, what is the the correct time? Thanks for help

First, will convert the minutes to decimal fractions:

1) 6:55 =6 + 55/60 =6.9167..... hours

2) 6:55 =6 + 55/60 =6.9167..... hours

3) 6:42 =6 + 42/60 =6.7            hours

4) 6:28 =6 + 28/60 =6.467....    hours

5) Since the professor set them an AVERAGE of 2 hours too slow, we therefore have:

6.9167 + 6.9167 + 6.7 + 6.467 =27

27/4 = 6.75 hours = 6:45 hours. But, since he set then 2 hours too slow, therefore we have:

6:45 + 2 = 8:45, which is the correct time.

Mmmm....there is no "trick'

V_sphere =  (4/3) * pi * (radius of the sphere)^3

-800 + 250  = -550  →  550 ft below sea level

Professor Freeman likes to to think that he has more time to finish his work than he really has. So, he has set the four clocks in his study to be an average of exactly 2 hours slow. If 2 of the professor's clocks read 6:55, a third reads 6:42 and the fourth reads 6:28, what is the the correct time? Thanks for help.

Let the elapsed minutes [since 12:00]  for the  4 clocks be   415, 415, 402 and 388  respectively

And the average of these minutes  =

[415 + 415 + 402 + 388] / 4  =  405 minutes

So......405 minutes after 12:00  =  360 + 45  =   [ 360 / 6] : [45mod 60]  =    6:45

So.......the correct time is actually 2 hours later =  8:45

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19

We can solve this as follows, where h is the height we are looking for.....we know the angle and the hypotenuse of a right triangle......and in both cases, we're looking for the opposite side....so the sine function will help us.....

sin 45  =  h/25  →   h  = 25sin45

And

sin 75  = h/25  → h =  25sin75

So.......we're looking for

25sin45 < h < 25 sin75   =  

17.68 ft < h < 24.15  ft     

If triangle ABC has AB=4; BC=5cm; AC=6; what is angle C?

Find the distance between the parallel lines m and n whose equations are y = x + 4 and y = x - 6, respectively.

There are several ways to do this...here's one

Let (0, 4) be a point on the first line

Then.......a line with a negative reciprocal slope going through this point will have the equation :

y = -x + 4........so......we can find the intersection of this line with y = x - 6....set both equations equal

-x + 4  = x - 6   add x, 6 to both sides

10 = 2x     divide both sides by 2

5  = x

So...using -x + 4, the y value  at intersection  = -1.......

So...we just need to find the distance from  (0,4)  to ( 5, -1)  =

√[ (5)^2 + (4 + 1)^2 ] =  5√2  ≈  7.07 units

Here's a pic....AB is the distance  with A = (0,4)  and B = (5, -1)

325+227=552

552/12=46

Take sqrt of both sides to get

+- sqrt(1/3) = t

+- 1/\(\sqrt{3}\)   =  t