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That is so unhelpful

Note that in every term, the numerator increases by 1, and the denominator decreases by 1. 

Because the first term starts in 1 and increases by 1, the number of terms is the last term. 

Can you take it from here?

Can I have a better explanation

(a) See the graph below.

(b) g(x) = 1/3*f(2x + 2).

(c) We stretch the graph horiztonally by a factor of 2, then stretch the graph vertically by a factor of 3, then shift down 2 units.

For how many real values of x is sqrt(120 - sqrt(x^2)) an integer?  

First note that sqrt(x²) is simply x.  But wait, it's not simply x, it is +x.  Aha. 

If you want to consider only the positive values of sqrt(x²), then there are these: 

sqrt (120 – 20)  =  10 

sqrt (120 – 39)  =    9 

sqrt (120 – 56)  =    8 

sqrt (120 – 71)  =    7 

sqrt (120 – 84)  =    6 

sqrt (120 – 95)  =    5 

sqrt (120 – 104)  =  4 

sqrt (120 – 111)  =  3 

sqrt (120 – 116)  =  2 

sqrt (120 – 119)  =  1              Looks like 10 of them.  If you count them this way.  

But if you want to take into account the negative root of x² then there is no finite number of values. 

For example, sqrt (120 – (–1))   =  sqrt (121)  =  11  

                     sqrt (120 – (–24))  =  sqrt (144)  =  12 

                     sqrt (120 – (–49))  =  sqrt (169)  =  13

As you can see, this will go on forever, so there is no numerical total. 

_.

This is a divisibility rule for 11. Here's more information about divisibility rules:

https://en.wikipedia.org/wiki/Divisibility_rule#:~:text=If%20the%20number%20of%20digits,must%20be%20divisible%20by%2011.&text=If%20the%20number%20of%20digits%20is%20odd%2C%20subtract%20the%20first,must%20be%20divisible%20by%2011.

Every other fraction has an even number on top and an even number on bottom, so those will reduce.  I don't know about the others.  

A broke the vase.

Yeah, there are \(1010\) of them, but the question asks for the number of the fractions expressed in lowest terms.

Note that in every term, the numerator increases by 1, and the denominator decreases by 1. 

Because the first term starts in 1 and increases by 1, the number of terms is the last term. 

Can you take it from here?

sorry, i meant lcm(b,c)

thank you so much!! i got the answer :D

5/4 wasn't right

Note that the area of the overlapping region is basically the area of the triangle without the 2 smaller triangles in the corner. 

Because the distance from the center vertice to the edge of the square is \(10 \div 2 = 5\), the smaller triangle has a height of 1. 

This means that the 2 smaller triangles are congruent and similar to the larger triangle by a ratio of \(1:6\) 

So, the area of each triangle is \({1 \over 6} ^2 \times 18= {{1 \over 2}}\), meaning the area of the overlapping region is \(18 - (2 \times {1 \over 2}) = \color{brown}\boxed{17}\)

Use the distance formula to calculate the distance between the 2 points (this is the radius)

Now, use the radius to find the area of the circle. 

There are \(4!\) ways to order the chickens 

There are \(6!\) ways to order the cats. 

There is only 1 way to order the dog. 

There are \(3!\) ways to order the chickens, cats, and dogs. 

So, there are \(4! \times 6! \times 3 = \color{brown}\boxed{51840}\) ways. 

7/2 isn't correct

1/2 (2week)=7  =>

1/2 (1week)= 7*2=14 =>

all levels = 14*2= 28

What is the smallest positive integer such that (98*k)^(1/3) is an integer

98 = 2 * 7^2

28 = 2^2 * 7

The smallest positive k == 28

98 * 28 ==2,744

2,744^(1/3) ==14

So, I got something for the first section!

There are three possible categories of arrangements for \((a, b, c)\). They are:

• \((a, b, c)\) are equal.
• \((a, b, c)\) are all different.
• Two variables are equal, one is not.

There are \(6\binom{n}{3}\) arrangements when all the variables are different. \(\binom{n}{3}\) accounts for all the arrangements regardless of order, but we must multiply by 6 to get our final answer.

There are \(n\) arrangements when all the variables are the same. 

There are \(3n(n - 1)\) ways to compute the arrangements when two variables are the same, and one is not equal.

Now, I need to figure out b; which I cannot.