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Because if we think about it slope is (Average velocity) is change in position / time traveled. 

I thought it might be "b" but wouldn't they normally put it over a fraction? That's what made me think Position "d"

Actually....I believe  "b"  is correct

Avg velocity  =    s(b)  -  s(a)

                          __________

                             b  -  a

So   

s(5)  -  s(4)

_________       =        s(5)  - s(4)

   5  -  4

V(t)  = 40(20 - t)^2 

V(t)  =  40 ( t^2 - 40t + 400)

V(t)  =  40t^2  - 1600t  + 16000

V ' (t)  =   80t - 1600

V' (10)   =  80(10) - 1600  =    -800 L / min

You were correct  !!!

Note, Julius, that the restriction is due to the fact that at 20 min....all the oil is drained from the tank.....

Thanks hectictar! 

@EP I think hectictar got it. Thanks anyways :)

\(\lim\limits_{ x \to 25^+ }f(x)\)  must be equal to \(\lim\limits_{ x \to 25^- }f(x)\)  so that the limit exists at the point with x-coordinate 25.

Where are the options, Julius?   

Ooops!  You left off the most important part of a)     

(c)  is the answer

Look at the graph of  1/ (x^2 + 1)   here :

https://www.desmos.com/calculator/zsurrlkbsc

Note that we have a horizontal asymptote of y  = 0, but there is no "break " in the graph !!!

 a⁴ b

_____

 c² d³

We can't really do anything here, Manuel.....we have all different variables and we don't know the value of any of them.....thus....there is nothing to "cancel" or "simplify"

Maybe it's a case of which fell off the log first....the chicken or the egg.....???

1.  Factor both expressions

x^2  +  x  -  12  =    (x + 4) (x - 3)

x^2 + 2x  - 15  =    (x + 5) (x - 3)

The LCM  is just all the different factors we have listed....so....

(x - 4)  (x + 5) ( x - 3)  ⇒ second answer

2.  Notice that we can see if the first numerator factors  using the second denominator

(w + 6) ( w - 4)                 8

____________  +         _____

(w + 6) ( w - 5)             ( w - 5)

( w - 4)       +     8

______          _____

(w - 5 )            (w - 5)

( w - 4 + 8)

_________

  ( w  - 5)  

w +  4

_____

w  - 5

3.   Here's a trick to always try on this kind of problem, NSS

Look at the second denominator.....see if the first actually factors with this

Notice that the first denominator is   ( b - 1) (b + 2)

So we have...also factoring the top of the first fraction

(b - 4) ( b + 2)                  6

___________    -          _____  

(b - 1) (b + 2)                 b -  1

(b - 4)                 6

_____      -      ____       

(b - 1)              (b - 1)

( b - 4  -  6 )

_________

    (b -  1)

b - 10

______

b  -  1

The trick here, NSS....is just to simplify the numerator first and then the denominator......so we have

2/5t  - 3/3t  =

2/5t  -  1/t          multiply the second fraction by  5 on top/bottom

2/5t - 5/5t   =   (2 - 5 )  / 5t   =   -3 / (5t)

The denominator becomes

1/2t  +  1/2t  =   2/2t  =  1/ (t)

So  we have

(-3/5)t  ÷ 1/t        flip the second fraction and multiply

-3 / (5t)  *  (t)/ 1  =    -3t / 5t   =   -3/5

Do you see how to do this??

Second one :

1/ 20   =  1/r₁  - 1/75       subtract  1/75  from  both sides

(1/20) - (1/75)  =  1/r₁

The common denominator  on the left is 300

15/300  - 4/300  =  1/r₁

(11/300)    =  1/r₁

To find r₁, just flip both fractions over

300/ 11  =  r₁/1

300/11  =  r₁ ≈  27 ohms

Nevermind, I managed to solve it. 

Opposite angles in a cyclic quadrilateral are supplementary....so....

x  + 4x  - 20  = 180     add 20 to each side

5x  = 200     divide both sides by 5

x  = 40°

So....angle B  =  (4(40) - 20)  =  160  - 20   =  140°

your very welcome...actually it's foghorn leghorn!!!....great to see a good sense of humour

whats the answer

The chicken !   No wait...... The egg!   No, wait.....

   Thanx for the clarification help !    

~EP

Perhaps the author of the question didn't believe it an issue otherwise they would have included that a-2b >= 0 in the question. I do see your dilemma but if you remember back to when you learnt all about logs and I guess it depends on how your teacher treated the topic. A logarithm as we know is the power of a base to obtain a given number. From this definition we get a whole new mathematics topic part of which is simplifications. The actual log law in question is that if we have a log of a power then :  log_bxⁿ  =  nlog_bx  not the other way round. Of course the other way has been adopted also to help with simplification. Maybe there are those who think its a case of which came first the chicken or the egg?

The second one    (the arc degrees intercepted by an inscribed angle is equal to TWICE the angle)

  The two angles cover the complete 360 degrees of the triangle in their arcs

soooo     360 = 2 ( 3x+1    +   2x+9  )

              180 = 5x +10

                 170 = 5x        x = 170/5 = 34  degrees

a = 2(34) + 9 = 77 degrees 

First one

Inscribed angles are 1/2 of the arc that they cover

  the 121 degree angle covers   121 x2 = 242 degreees of the circle......

   the other angle covers  360 - 242 = 118 degrees 

     so the angle in question = 118/2 = 59 degrees

First one (the rest are very similar)

3i       3 (1)  + 3(2)  + 3 (3)  + 3(4) +3 (5) =

             3          6          9          12     15     = 45

Think you can try the others now that you see how to do it?

The second one is the second choice

when j =

1 the value is       -5

2                          -1

3                           3

4                           7

5                         11

I think the first one is the 4th choice   

j =

1        1

2        1/4

3        1/16

4         1/64

5        1/256       (I think you typed the red terms incorrectly in your question.