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Where is "as shown below", lazy ?

The answer is a = 4, b = 2, c = 2.

Thanks, that's right!

The slope of the line is is (2 - 1.2)/(5 - 1) = 1/4.

Then by point slope form, the equation of the line is

y - 2 = 1/4(x - 5)

y - 2 = 1/4*x - 5/4

y = 1/4*x + 3/4

Did you subtract one number from the other and see the pattern?

30  -  16 ==14

44  -  30 ==14

58  -  44 ==14.......what is the pattern?

log(3, 1000) ==log(1000) / log(3) ==3 / 0.47712125==6.2877 ==6 (rounded down - which is the same as its floor)

By change of base, log_3 1000 = log(1000)/log(3) = 3/1.0986... = 2.730...

Taking the floor, we get an answer of 2. 

The perimeter of rectangle OSTQ is 22.

a=listfor(n, 0, 8, (12 - n) nCr 4);printa, "==", sum a ==(495, 330, 210, 126, 70, 35, 15, 5, 1) == 1,287 ways.

OR:

[12 C 4] + [11 C 4] + [10 C 4] + [9 C 4] + [8 C 4] + [7 C 4] + [6 C 4] + [5 C 4] + [4 C 4] ==495 + 330 + 210 + 126 + 70 + 35 + 15 + 5 + 1==1,287 ways.

I think answer 1 is the correct one.  :)

Yes it is a sloppy question but you are basically expected to assume that it is the height of the bird that you are finding and that the ground is all level.

The answer I got was 39 feet.   Do you want me to explain further or did you just want to check your answer.?

Franklin can take a step to the right, left, up, or down at each point. Therefore, each step corresponds to one of four directions.

After 13 steps, there are a total of $4^{13}$ different sequences of steps that Franklin can take.

However, we need to determine the number of different points that Franklin could end up at. To do this, we can consider the net displacement of Franklin's position. Let's denote right as $(1,0)$, left as $(-1,0)$, up as $(0,1)$, and down as $(0,-1)$.

For Franklin to end up at a specific point $(x, y)$, he must take $x$ steps to the right and $y$ steps up, with the remaining steps taken to the left and down. The total number of such sequences of steps is given by the binomial coefficient $\binom{13}{x}$ for each $x$ from $0$ to $13$.

Similarly, we can calculate the number of sequences for the number of steps taken in the left, down, and up directions.

Therefore, the number of different points that Franklin could end up at is:

\(\sum_{x = 0}^{13} \sum_{y = 0}^{13 - x} \sum_{z = 0}^{13 - x - y} \binom{13 - x - y}{z}\)

Calculating this sum yields a result of 214 paths.

To find the number of different sequences of rolls that could have resulted in a product of 900, we need to consider the prime factorization of 900.

The prime factorization of 900 is:

900 = 2² * 3² * 5²

We can distribute these prime factors among the five rolls of the die. Since the order of the rolls matters, we can think of each roll as a "slot" where we can place one of the prime factors.

Let's consider the number of ways to distribute the prime factor 2² = 4 among the five rolls. We can have:

4 in the first roll, 0 in the second, third, fourth, and fifth rolls.

3 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.

2 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.

2 in the first roll, 1 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

1 in the first roll, 1 in the second, 1 in the third, and 2 in the fourth and fifth rolls.

Now, let's consider the number of ways to distribute the prime factor 3² = 9 among the remaining slots. We can have:

9 in the first roll, 0 in the second, third, fourth, and fifth rolls.

8 in the first roll, 1 in the second, 0 in the third, fourth, and fifth rolls.

7 in the first roll, 2 in the second and 0 in the third, fourth, and fifth rolls.

6 in the first roll, 3 in the second, 0 in the third, fourth, and fifth rolls.

6 in the first roll, 2 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

5 in the first roll, 4 in the second, 0 in the third, fourth, and fifth rolls.

5 in the first roll, 3 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

4 in the first roll, 4 in the second, 1 in the third, and 0 in the fourth and fifth rolls.

4 in the first roll, 3 in the second, 2 in the third, and 0 in the fourth and fifth rolls.

3 in the first roll, 3 in the second, 2 in the third, and 1 in the fourth and fifth rolls.

Finally, for the remaining prime factor 5² = 25, we have only one possibility:

25 in the first roll, 0 in the second, third, fourth, and fifth rolls.

By multiplying the number of possibilities for each prime factor, we find the total number of different sequences of rolls:

5 * 10 * 1 = 50

Therefore, there are 50 different sequences of rolls that could have resulted in a product of 900.

What do you expect the answer to look like ?

Would you regard

\(|z-(5-2i)|\leq4,\)

for example as an interval ?

I also commented on the original :)

Nicely answered guest. :)     

My only suggestion is that you could have simplied the first equation to make it a lettle simpler

(10A + B)  =  (10B + A) + 45

9A-9B=45

A-B=5

But that is certainly no big deal :)

A sloppy question.

To begin with we need to know whether the 10ft away from the tree was on level ground or up or down a slope.

Further we are not told where the bird was in the tree, at the top or further down in the branches, or are we supposed to assume ?

Tighten up on the details and we may be able to do something.

There are 6 variables

One could be 8 and the rest zero        6 ways

One could be 7,1,0,0,0,0                  6*5=30 ways

6,1,1,0,0,0                                  6*5C2 = 60 ways

6,2,0,0,0,0                                  6*5 = 30 ways

5,1,1,1,0,0                                 6*5C3 = 60 ways

5,2,1,0,0,0                                 6*5*4 = 120 ways

5,3,0,0,0,0                                6*5 = 30 ways

4,1,1,1,1,0                                6*5 = 30 ways

4,2,1,1,0,0                                 6*5*4C2 = 180 ways

4,2,2,0,0,0                                6*5C2 = 60 ways

4,3,1,0,0,0                                6*5*4 = 120 ways

4,4,0,0,0,0,                                6*5 = 30 ways

3,1,1,1,1,1                                 6 ways

3,2,1,1,1,0                                 6*5*4 =  120 ways

3,2,2,1,0,0                                 6*5*4C2 = 180 ways

3,3,1,1,0,0                                 6C2*4C2 = 90 ways

3,3,2,0,0,0                                 6C2*4 = 60 ways

2,2,2,2,1,1                                 6C2=15 ways

2,2,2,2,2,0                                 6 ways

Add all those up and I think you will have your answer. :)                  

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Let the number = N

5N / (N + 7) = 0, solve for N

Cross multiply

5N = 0

N = 0

The diagonals of a square are equal and bisect each other at right angles. 

Therefore, angle APB is a right angle and the triangle APB is an isosceles right triangle with sides of 1.

By Pythagoras' Theorem:

x^2 = 1^2  +  1^2

x^2 = 2

x =sqrt(2) = 1.41421.....

There are 3 combinations as follows:

(1, 5, 5, 6, 6) ==5!/2!2! ==30 permutations 
(2, 3, 5, 5, 6) ==5!/2!    == 60 permutations  
(3, 3, 4, 5, 5) ==5! /2!2! ==30 permutations

Total ==30 + 60 + 30 == 120 different sequences of rolls.