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floor [ (2 -3x) / (3x + 8) ]

The value  of  f(1)  to f(1000)  always =   -1

So....the sum is  -1000 

Area =  6 (1/2) * 3^2 * sqrt (3) / 2  =

27 sqrt (3) / 2 ≈  23.38

I think there was a slight error in your code for python. 

I followed your logic and iteration, and the answer turned out to be 300. 

I don't know if I made an error, but it does match with the answer I got using plain logic. 

Here's my explenation for the problem. 

First, let's find how many of each number there are. 

500/3 = 166.66, so there are 166 integers. 

500/4 = 125, so there are 125 integers.

500/5 = 100, so there are 100 integers. 

Next, we must find the duplicates of numbers. 

3*4 = 12 

3*5 = 15

4*5 = 20

500/12 = 41.67, so there are 41 integers

500/15 = 33.33, so there are 33 itnegers. 

500/20 = 25, so there are 25 integers. 

LCM = 3*4*5 = 60. 

500/60 = 8.33, so there are 8 integers. 

166+125+100-41-33-25+8 = 300. 

Let me summarize. 

First, we find how many numbers are divisble by 3,4,5. However, there are duplicates, so we subtract those, before adding the LCM!

Thanks! :)

First, let's find how many of each number there are. 

500/3 = 166.66, so there are 166 integers. 

500/4 = 125, so there are 125 integers.

500/5 = 100, so there are 100 integers. 

Next, we must find the duplicates of numbers. 

3*4 = 12 

3*5 = 15

4*5 = 20

500/12 = 41.67, so there are 41 integers

500/15 = 33.33, so there are 33 itnegers. 

500/20 = 25, so there are 25 integers. 

LCM = 3*4*5 = 60. 

500/60 = 8.33, so there are 8 integers. 

166+125+100-41-33-25+8 = 300. 

Let me summarize. 

First, we find how many numbers are divisble by 3,4,5. However, there are duplicates, so we subtract those, before adding the LCM!

Thanks! :)

Let's say there are x sides on the polygon. 

The number of diagonals can be represented by \( x*(x-1)/2 - x \), meaning we can write the equation

\(2x = x*(x-1)/2 - x\). 

Solving for this equation, we find that x = 7, which is a heptagon. 

A heptagon has 7 sides and indeed has 14 diagonals!

Thanks! :)

To find the possible sets, let's call the integers a, b, c, d, and e, listed from least to greatest. We know that the mean, median, and range are all 7.

The mean of the set is the sum of the numbers divided by their count. So, we have a + b + c + d + e = 5 * 7 = 35.

Since the median is the middle number when the set is sorted, the third number must be 7. So, we have a, b, 7, d, and e.

The range is the difference between the greatest and least numbers. Since the range is 7, we have e - a = 7.

Now, since e > a, we know that e is at least 8. This means the maximum value for e is 15, since e + 7 = a + 14.

To find the number of distant sets, we need to consider all possible values for e.

If e = 8, then a = 1. The set would be {1, b, 7, d, 8}.

If e = 9, then a = 2. The set would be {2, b, 7, d, 9}.

Continuing this pattern, if e = 15, then a = 8. The set would be {8, b, 7, d, 15}.

So, we have a total of 15 - 8 + 1 = 8 distant sets whose members are listed from least to greatest and have the given properties.

Area of a hexagon =  6* (1/2) s^2* sqrt (3)/2      where s is  the side length

So

3 = 6 * (1/2) * s^2 * sqrt (3) / 2

3 = 3*  s^2 * sqrt (3) / 2

s^2 = 2/sqrt (3)

s = sqrt [ 2 / sqrt 3] ≈ 1.075

Note that the sum of the first n terms =   n * a1  +  d * (n)(n-1) / 2     where n > 1

So

S10  =  10a1 + d (10)(9) / 2  =   10a1 + 45d  =  0         (1)

S20 =   20a1 + d(20)(19) / 2  =   20a1 + 190d = 1/5     (2)

Multiply (1) by -2  and add to (2)  giving

100d = 1/5

d = 1/500

To find  a1

10a1 + 45 (1/500)  = 0

10a1 = -45/500

a1 = -45 / 5000  =  -9 / 1000

So

S70  =   70 (-9/1000) +  (1/500) (70)(69) / 2  =    21 / 5  =  4.2

log₂ (2x)  = log₄ (16 + x)

Using the change-of-base theorem

log (2x) / log2  = log (16 + x) / log 4

log (2x) / log 2 = log (16 + x) / log 2^2

log (2x) / log 2  = log (16 + x) / [ 2 log 2]      multiply through by log 2

log (2x)  = log (16 + x) / 2

2log (2x) = log (16 + x)

log (2x)^2 = log (16 + x)      this implies that

(2x)^2 = 16 + x

4x^2 - x - 16  =  0

Using the Quadratic Formula  (and taking the positive value of x )

x =  [ 1 + sqrt [ 1^2 - 4 (4) (-16) ] ] / (2 *4)

x = [ 1 + sqrt (257) ] / 8

Using combination, the number of ways of placing 8 counters in the square is 12 ways and the number of ways of placing 12 counters in the square is 8 ways

What is the number of ways of placing 8 counters in the square

(a) To place 8 counters in the 4 x 4 grid, with exactly two counters in each row, we can first choose the two columns in the first row that will have counters. Using combination, there are {4 C 2} = 6 ways to do this. Then, we choose the two columns in the second row that will have counters, but we need to make sure that these columns are not the same as the ones chosen in the first row. There are only two columns left to choose from, so there are only two ways to do this. Similarly, we choose the two columns in the third row that will have counters, making sure that they are different from the columns chosen in the first two rows. There is only one column left to choose from, so there is only one way to do this. Finally, the columns for the fourth row are uniquely determined. Therefore, the total number of ways to place 8 counters in the grid with exactly two counters in each row is:

6 * 2 = 12

(b) To place 12 counters in the 4 x 4 grid, with exactly three counters in each column, we can start by choosing the three rows that will have counters in the first column. Using combination, there are {4 C3} = 4 ways to do this. Then, we choose the three rows that will have counters in the second column, but we need to make sure that these rows are not the same as the ones chosen for the first column. There are only two rows left to choose from, so there are only two ways to do this. Similarly, we choose the three rows that will have counters in the third column, making sure that they are different from the rows chosen for the first two columns. There is only one row left to choose from, so there is only one way to do this. Finally, the rows for the fourth column are uniquely determined. Therefore, the total number of ways to place 12 counters in the grid with exactly three counters in each column is:

4 * 2 = 8

Thus, there are 12 ways to place 8 counters in the grid with exactly two counters in each row, and 8 ways to place 12 counters in the grid with exactly three counters in each column.

Thanks! :)

We can use constructive counting to solve this problem. 

CASE 1

First, let's see all the two digit numbers that satisfy these conditions. 

We only have two, with 23 and 32. 

CASE 2

Now, let's check all the 3 digit numbers. 

We have 311, 131, 113, giving us 3. 

We also have 221, 212, and 122, giving us another 3. 

Case 3

4 digit number time! This is where it gets a bit complicated, but it's still not that bad. 

We have numbers with three 1s and one 2, which has 4 different possibilities. 

Case 4

Last case ! 5 digit numbers only have 1, with 11111. 

Now, we just have to add up all our cases, and we get 2+3+3+4+1 = 13. 

We have 13 numbers that satisfy the conditions. 

NOTICE: This problem didn't have a lot of cases, so it would be easy to count. Bigger and wider ranges would require permutations!

Thanks! :)

First off, we can subsitue y out to get \(\log _4\left(16+x\right)=\log _2\left(2x\right)\). 

Now, we just have to solve for x. \(\log _4\left(16+x\right)=\log _2\left(x\right)+1\). 

I'm not sure exactly how to explain, but now we just have to isolate x and we eventually get

\(x=\frac{1+\sqrt{257}}{8}\)

Here's a graph to make it more clear.

Their intersection point is our answer for x! 

Thanks! :)

All points 5 units away from (2, 7) means that we can have a circle with radius 5 and center (2, 7). 

Such a circle would look like \((x - 2)^2 + (y -7)^2 = 25 \). 

Now we just have to find the intersection points of this circle and \(y = 5x - 28\). 

This is pretty simple. We just have to find ordered pairs (x, y) that match for both equations. Luckily, we already isolated y from the second equation, so plugging in that y value into the first equation, we have

\((x -2)^2 + (5x -28 - 7)^2 = 25 \\ (x - 2)^2 + (5x - 35)^2 = 25\)

\(x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25 \\ 26x^2 - 354x + 1204 = 0\)

Solving from this, we get \(x=86/13, 7\). 

Now plugging these values back into our second equation, we get \(y = 66/13, 7\). 

So our two points are (86/13, 66/13) and (7, 7)! 

Thanks! :)

What real value of $t$ produces the smallest value of the quadratic $t^2 -9t - 36 + 13t - 60$?  

Combine like terms                                  t² + 4t – 96   

Arbitrarily say this equals "y"            y  =  t² + 4t – 96   

The "degree" of an equation is determined by the highest power of a variable in the equation. 

Since the highest power is the squared term, we know the curve is a parabola that opens upward.  

The smallest value of "y" is the bottom where it turns around and goes back up.  

We can find this by taking the first derivitive of the equation and setting it equal to zero.  

The first derivitive defines the slope of a curve at any point. 

So the first deritive of t² + 4t – 96 is 2t + 4  

Set equal to zero is                           2t + 4  =  0   

                                                               2t  =  –4   

                                                                 t  =  –2   

I took the equation to www.desmos.com/calculator to verify the answer.  

_.

First off, let's simplify and combine all like terms. 

We get \(t^2+4t-96\). Now, we have a perfect parabola we can graph! The vetex of the graph is the minimum value point, and taking the x value from that coordinate pair will get us t. 

The vertex we get is (-2, -100), which is our minimum point. 

So t=-2 is our answer!

Thanks! :)

This problem isn't too bad to solve!

First off, we need an equation to figrue out the 3 new numbers we achieve. Let's let n be the constant we add. 

Since all terms of a geomtetric sequence have equal fractions, we can write the equation

\(\frac{120+n}{60+n}=\frac{160+n}{120+n}\\ (120+n)^2=(60+n)(160+n)\\ 14400+n^2+240n=9600+n^2+220n\\ 4800=-20n\\ n=-240\)

I know the constant is now negative, but you do get the numbers -180, -120, and -80, which do in fact form common ratios. 

-180/-120 = 3/2. So 3/2 is the answer. 

I'm not sure if this was the exact answer you were looking for, but it does work

Thanks! :)

We can write

a1 + (a1 + 2d) + (a1 + 4d) + (a1 + 6d) + (a1 + 8d) + (a1 + 10d) =  6a1 + 30d = 0

(a1 + d) + (a1 + 3d) + (a1 + 5d) + (a1+ 7d)+ (a1 + 9d) + (a1 + 11d) = 6a1 + 36d = 0

So

6a1 + 30d  = 0   

6a1 + 36d  = 0

6a1 + 30d = 0

-6a1 - 36d = 0        add these

-6d = 0 

d = 0

So

6a1 + 30(0) = 0

6a1 = 0

a1  = 0

Laverne starts counting out loud by ${}5$'s.  She starts with $2$.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds $20,$ Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running?  

Laverne's           Shirley's  

Counting          Summation  

     2                           2  

     7                           9   

   12                         21    

_.

Simplify  as

-5s^2  -16s + 39

This is a parabola that opens downward.....there is  no minimum  value

https://web2.0calc.com/questions/sum_93

The psychology textbooks can be grouped together in one of  five positions....and  for each of these, they can be arranged 3!  = 6 ways

And for each of  the above arrangements the math textbooks can be  arranged in 4! = 24 ways

So

5 * 3! * 4!   =

5 * 6 * 24  =

720 ways

Combining the like terms, we get \(4t^2+4t+14\). 

Here's the thing though. The graph of this doesn't actually have a maximum. The parabola is facing upwards, meaning the maximum is infinite!

The value of t that would get us the minimum value would be -0.5!

Thanks! :)