All Answers 355923 Answers

I would show all the steps to the problem, but that would be a bit time consuming, so I will highlight the way to go. 

We just have to use the distributive property to get \({x^{3}\left(x^{4}-8x^{3}+17x^{2}-23x+14\right)+x^{2}\left(x^{4}-8x^{3}+17x^{2}-23x+14\right)+x\left(x^{4}-8x^{3}+17x^{2}-23x+14\right)+1\left(x^{4}-8x^{3}+17x^{2}-23x+14\right)}\)

Eventually, we get \(x^{7}-7x^{6}+10x^{5}-13x^{4}+8x^{2}-9x+14 \). 

So the coefficient of x is -9. 

Thanks! :)

First, let's multiply the two terms together. 

We have \(x^3(x^2+tx)\) = \(x^5 + x^4t\). 

In general, I think t could be any number as long as it doesn't have an x with power above 21. 

I hope I answered your question!

Thanks! :)

According to the info given, Each additional worker hired reduces the time by 4 days

We just have \(4 x = 20 \) where x is the number of workers needed. 

This means we have x = 5, so we need 5 additional workers!

Thanks! :)

https://web2.0calc.com/questions/triangle_49107

When Dirk finishes, Edith has run 14km  and Foley has run 9 km

This means that Foley runs 9/14 as fast as Edith

When Edith has run 24km, Foley has run \( 24 (9/14)  ≈  15.43  km\)

So Foley is \(  24 - 15.43  ≈  8.57\) km behind!

Thanks! :)

The formula for the area of the parralleogram is \(base*height\), just like a rectangle! 

Thanks! :)

Since we know AD bisects BAC, we have

BD  = 3 - x

CD =  x

Now, we can write

\(BD / AB = CD / AC \\ (3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

This just gives us

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3\)

Thanks! :)

So we know that

\(CD = BD = 10 \\ ED = ED\)

Through this, we can find that triangle CDE is congruent to triangle BDE.

This means BE = CE!

\(sin CED / CD = sin CDE / CE \\ sin 75 / 10 = sin 90 / CE \\ CE = 10 sin 75 = BE ≈ 10.35\)

Thanks! :)

The number of students ONLY in math class is a-b. 

The number of students ONLY in science class is c-(a-b) = c-a+b. 

The number of students ONLY in science class is just c-a. 

So, we just have a-b + c - a = c - b. 

Thanks! :)

The freshmen are able to stand in 4! = 24 ways. There are 4 ways we can put them into the line with them all standing together. 

The sohphomores can staned in 3! = 6. 

We just have to do 24*4*6 = 576. 

There are 576 ways we could arrange them!

Thanks! :)

(b) For question b, we basically do the same exact thing as stated above. 

First, let's review the cases for \(a=b=c \). 

There are obviously n such triplets. 

Next, let's look at when exactly two are the same. You have 3 choices for which number is repeated and then n choices for the repeated number, and then n-1 for the distinct number. This gets us 3n(n-1). 

Lastly, we have 3 different numbers. There are \(6\)\(n \choose 3\) ways because there are 3 distinct numbers out of n and 3! = 6 ways to do this. 

Thus, when we add them together, we get \(n + 3n(n - 1) + 6 \binom{n}{3}.\)! 

Thanks! :)

First, let's see how many ways there are to arrange the 3 psychology books. 

We have 4p3 ways the arrange the books multiplied by the number of ways to put these books into the group of 7, which is 4. 

We have 24*4 = 96. 

Now, there are 4p4 ways to arrange the math books, which is simply 24. 

96*24 = 2304. 

There are 2304 ways to arrange the books! 

Thanks! :)

We would just have \(3!*4! = 6*24= 144\) ways, since we must place a children down before completing the problem. 

Thanks! :)

Alright, I'll try to tackle this problem!

(a) We can try to count the number of ordered pairs (a, b) ourselves. We can split these into two cases. 

1. \(a=b\)

2. \(a \neq b\)

1. Since we can choose any numbers between 1 and n, there are obviously n possibilities we can achieve. 

2. Since we can choose any 2 numbers between 1 and n that are not equal, we just have \(n \choose 2\). 

Hence, we just have \(n+\)\(n \choose 2\) \(+ \)\(n \choose 2\), which is exactly what we wanted in the first place. 

I hope I answered part (a) correctly!

Thanks! :)

Let's let the two sides of the rectange x and y. 

We can use the problem to write the equation

\(2x + 2y = 40 \\ x + y = 20 \)

Isolating y, we have \(y = 20 - x\). 

Subsituting this value for y into the second part of the qeustion and using the pythaogrean theorem, we have

\(x^2 + y^2 = (10 \sqrt{2}) ^2 \\ x^2 + ( 20 - x)^2 = 200 \\ x^2 + x^2 - 40x + 400 = 200 \\ 2x^2 - 40x + 200 = 0 \\ x^2 - 20x + 100 = 0\)

Finally, we find that we have x = 10. From the first equation, we find that y = 10 as well. 

The area is just 10*10=100!

Thanks! :)

This isn't too hard of a question once you realize how to do it!

\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ [{EFGH}]=\Delta EFG\cdot \frac{1}{3}h\\ [{EFGH}]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [{EFGH}]=9\)

The key is realizing that the pyramid is just 1/2 of EFG times the height!

Thanks! :)

This isn't too bad of a problem!

First off, we have to find how many numbers between 1 and 1000 are divible by 5. 

It might seem challenging, but what it really is is just 1000/5 = 200! 

Now, there are 1000 numbers between 1 and 1000, so we just have 200/1000, which gets us 1/5. 

So there is just a 1/5 chance that there will be an integer divisble by 5!

Thanks! :)

First, we have to write this in standard form by combining all like terms and moving all terms to one side. We get \(x^2+11x-49>0\)

Using the quadratic equation, we get \(x<\frac{-\sqrt{317}-11}{2}\quad \mathrm{or}\quad \:x>\frac{\sqrt{317}-11}{2}\). 

Putting this into interval notation, we have \((\frac{-\sqrt{317}-11}{2}, \frac{\sqrt{317}-11}{2})\). 

This isn't a hard problem to do. It is annoying we have square roots and an unfactorable polynomial, but not too bad in general!

Thanks! :)

The polynomials are 2x^2 + 4x and 4x^2 + 8x.

Yes, there are numbers that do not go through the loop in the described sequence of rules. This sequence defines a function that iterates on a number based on some rules. The loop you observed happens because the function eventually reaches numbers that have already been seen in the sequence.

Here's why some numbers might not enter the loop:

Odd numbers greater than 1: The function multiplies odd numbers by 3 and adds 1. If the resulting number is even (divisible by 2), the loop continues. However, if the resulting number is odd and greater than 1, the function will repeat the multiplication by 3 and addition of 1. This can potentially lead the sequence to entirely new, unseen numbers that haven't been encountered before.

Numbers divisible by 3 but not by 6: When a number is divided by 2, it might reach a point where the result is divisible by 3. If this number is divisible by 3 but not by 6 (meaning it has a remainder of 3 when divided by 6), the function will reach a number that is odd and greater than 1 (as described in point 1). This can again lead the sequence to explore new parts that haven't been seen before.

For example, the number 10 never enters the loop you observed. Here's why:

10 -> 5 (even, divided by 2) 5 -> 16 (even, divided by 2) 16 -> 8 (even, divided by 2) 8 -> 4 (even, divided by 2) 4 -> 2 (even, divided by 2) 2 -> 1 (odd, less than or equal to 1) - The sequence reaches 1, which terminates the process.

In conclusion, while the loop you observed occurs for many numbers, there are indeed numbers that the function explores and never enters the repetitive cycle. These numbers can fall under the categories mentioned above (odd numbers greater than 1 and certain numbers divisible by 3 but not by 6).

First, let's find how many ways we can choose 3 randomly, no restrictions. 

Using the glorious formula for combinations, we have \(9 \choose 3\)\(= 84\). This means there are 84 ways to choose 3 circles randomly. 

Now, let's see how many DON'T work. 

There are 6 rows and columns of 3 circles, and each of these adjacent 3 have 3 ways we can have an inablid combination. 

Thus, we have 6*3 = 18. 

\(84-18=66\). There are 66 ways we can do this!

I'm not really sure if this is correct...

Thanks! :)

I graphed it using a graphing calculator by graphing each side of the equation

The intersection gets us something like\(x \approx 11.18119077\)

I don't think this was the exact answer you were looking for, but my brain isn't smart enough to solve this ;)

Thanks! :) 

Alpha  =  the measure of the interior angle of a 15-gon  (a pentadecagon) =

[ (n - 2) * 180 ]  / n  =    [ (15 - 2) * 180 ]  / 15  =  156°