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Thanks   CalTheGreat, 

I could not remember who posted it otherwise I would have done that.

Also, I did not remember that I had responded or else I might have found it that way.

100% x    + 50% (5)   = 80% (x+5)

1.0x   +.5(5) = .8(x+5)        solve for x   the number of gallons of 100% acid to use,....

Sorry  that was    4 ways with free space    out of 12 ways    4/12 =   1/3

there are 5 ways to bingo across      1 has free space

2 diagonal ways    both have free spce

5 vertical ways....one has free space

4 with space   out of  17       4/17

Hey there guest! Glad to see you posting something about logarithms!(I need to brush up on them lol).

Let's get to the problem.

It's first given that f(n) is the base 10 logarithm of the sum of the elements in the nth row of pascals triangle. 

A key thing that you should realize is that the nth row of pascals triangle can be represented as: \({n \choose 0} + {n \choose 1} + {n \choose 2}.... {n \choose n}\)

Coincidentally, this also equals \(2^n\). You can try this out yourself with the first few rows of pascal's triangle!

The function f(n) then becomes:

\(f(n) = \log_{10}2^{n}\)

The expression we are asked to simplify then turns into:

\(\log_{10}(2^n)/ \log _{10} 2\)

Here, we have a neat log property we can use:

\(\log_{a} b / \log_{a} c = \log_{c} b\)

We then convert

\(\log_{10}(2^n)/ \log _{10} 2 = \log_{2}2^n\)

With the basic definition of a log, we then have

\(\log_{2}2^n = n\)

So the expression in terms of n is just n

Quick edit / note : Remember that the "first row" of pascal's triangle with just 1 term, is actually counted as the "0th" row.

In triangle ABC, we have AB=3 and AC=4. Side BC and the median from A to BC are of equal length, as in the picture below:

Hi booksly002, in order to find out angle A, you need to calculate side BC first.

Side  BC ≈ 3.163    

Thank you so much! That makes a lot of sense.

First, we start with placing the 4x4 square so that one corner is (0,0), and the opposite corner is (4,4). We can define the x and y "coordinates" of Larry's square as x₁ and y₁. We define for laura's square x₂ and y₂ for the coordinates. That means that Larry's square and Laura's square will only ever intersect when the intervals of X1 and X2 intersect, and likewise for Y1 and Y2. 

Let X₁ equal to [a, a+1] and X₂ equal [b, b+1] with \(0 \leq a \leq3\) and \(0 \leq b \leq3\). This implies that the intervals given(x1 and x2) would only intersect in the case where |a - b| = 1(since that means the distance would be less than 1, which is the length of one of the unit squares. The region where the following holds true would be partitioned off into two 45-45 90 triangles, both with leg length of 2. The area of the desired region would be:

\(9-(2*2*1/2)*2 = 9-4= 5\)

Now, we have the probability that the intervals X₁ and X₂ intersect, which is \(5/9\). Likewise, the probability that Y₁ and Y₂ intersect is also \(5/9\), so our answer is just:

\((5/9)^2 = 25/81\)

I feel like I've done this problem somewhere before lol, can't seem to remember where

n * n!  +  n!   =  720

Note  that if n = 5  then

5 * 5!  + 5!   =

5 (120)  + 120  =

600  + 120  =   

720 !!!

I got the answer

Consider the polynomials f(x)=4x^3+3x^2+2x+1 and g(x)=3-4x+5x^2-6x^3. Find c such that the polynomial f(x)+cg(x) has degree 2.

Rewrite  g(x)  as

(-6x^3 + 5x^2  - 4x  +  3)

Note  that  if  f(x)  + cg(x)   results  in a polynomial of degree 2.......we  need to find c  such that

4x^3  + c( -6x^3)  =    0

So....this implies  that

4  - 6c   =   0

4   =  6c

c =  4/6  =   2/3        

any time!

This is an upward -turning parabola

In  the  form  y  = ax^2 + bx + c.....the  x  value  that will give the minimum  is   -b / [2a]

So  we have  that     -b  =  12    and  a  =  2

So...the minium occurs  at  x  =    12 / [2(2) ] =  12 / 4  =   3

Thank You!

LOL!!!!!!! Anyway, here's another way to do it.

x=Plane's rate in still air

y= Rate of the wind

x-y=550

x+y=870     

Then, use the elimination method. (subtract)

-2y=-320

y=160km/h

Then, we know that the plane's rate in still air is 550+160, which is 710 km/h.

It is DEFINITELY A STRONG WIND!!!!!!!

This was also posted here: https://web2.0calc.com/questions/4-let-f-x-1-x-be-the-parent-function-let-g-x-3-x

Take a look at it again!

THX, CTG   !!!

I agree!

We can just factor it into (n-7)(n+2).

n=7

n=-2

So the smallest value that will make this function true is -1.

Nice job, Chris!!!

n^2  - 5n  - 14 <  0     factor

(n - 7) ( n + 2)  <  0

The  function will be negative  on  the interval   ( -2, 7)

So.....the smallest integer  that  makes the function  negative  is   n    =  -1

GM....I don't  know  (b)  or (c)   but  here's (a) and (d)

(a)  

             3

x - 4  [ 3x   - 10   ]

           3x   - 12

          __________

                       2

So  g(x)  =   3  +   2 / ( x - 4)

(d)

Using  the first function....the  horizontal asymptote   is

y  =  3 / 1      =  3

And the vertical symptote    is  x   = 4

Flying against the wind, an airplane travels 3850 km in 7 hours. Flying with the wind, the same plane travels 4350 km in 5 hours. What is the rate of the plane in still air and what is the rate of the wind?

Let P  be the plane's rate in still air

Let the rate  of the wind   = W

So  we  have  that  Time *  Rate   = Distance

So

7(P - W)  = 3850     divide  both sides  by  7

5(P + W)  =  4350   divide both sides  by 5

P - W   =  550

P + W =  870      add these

2P  = 1420     divide both sides  by 2

P  =  710  (kph)

And

710  - W =  550

W =  710 - 550

W = 160 [kph]

That's a STRONG WIND   !!!!

2log₂ ( x + 5)  =  log₂(x - 9) + log₂(x + 53)  +  1

Note  that  1  =  log₂ 2

So  we  have

2log₂ ( x + 5)  =  log₂(x - 9) + log₂(x + 53)  +   log₂ 2

By a couple of log properties we  can write

log₂ ( x + 5)^2  =  log₂  [ (x - 9)(x + 53)(2) ]

The log base is the same so we can solve  this

(x + 5)^2  = (x - 9) ( x + 53)(2)           simpify

x^2 + 10x + 25  =  (x^2  + 53x - 9x  - 477) (2)

x^2  + 10 + 25  =  (x^2  + 44x  - 477) (2)

x^2  + 10x +  25   = 2x^2  + 88x  - 954    rearrange as

x^2 +  78x  - 979   =  0        this can  be  factored as

(x  -11) ( x + 89)   =   0

Setting both factors to 0  and  solving for  x   gives  us

x  = 11    or  x   = -89

The second answer gives us the log of a negative  number in the original equation....so..... it's no good

The only answer is

x  = 11

2. Let a and b be the solutions to 7x^2 + x - 5 = 0. Compute (a - 4)(b - 4).

Again, by Vieta

a + b  =  -1/7      

ab   =  -5/7

So

(a - 4) ( b - 4)   =

ab  - 4a - 4b  + 16   = 

ab  -4 (a + b)  + 16    =

-5/7  - 4(-1/7)  + 16   =

-1/7  +  16  =

-1/7  +  112/7  =

111 / 7