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Hmm..wonder why my picture was blocked....

Find the area of this shape.

Hello Guest!

\(A=16cm\times 7cm+\frac{1}{2}\times (16-8)cm\times (13-7)cm\)

\(A=136\ cm^2\)

  !

agreed 

I'm pretty sure this is fake as the real Richard wouldn't use "plz" and an emoji...he'd probably type more professionally.

Katie has a collection of nickels, dimes and quarters with a total value of $3.95. There are 4 more dimes than nickels and 3 more quarters than nickels, how many of each coin does she have?

Hello Katie!

\(5n+10d+25q=395\\ d=n+4\\ n+3=q\\ \)

\(5n+10(n+4)+25(n+3)=395\\ 5n+10n+40+25n+75=395\\ 40n=280\)

\(n=7\\ q=10\\ d=11\)

\(Hi \ Katie,\ you\ have\ seven\ nickels, ten\ quarters\ and\ eleven\ dimes.\)

  !

Smart!

You shall be awarded "all real numbers"

Not very useful in a battle, though... You'd probably lose against my almighty square root...

:)

Ahhh. Yes! Let me correct my mistake, and also I will give you some credit!

Hello ilorty.  That's a very good analysis.  One slip: "After 4 hours: 100 : 50"   Actually, after 4 hours Liz is starting at 0.  But that isn't important.  I figured you would know the flaw in the "half distance" reasoning; I was trying to prompt the OP (original poster) to think about it. 

_.

Very smart, "clairvoyant"

I'm sorry if I sounded a bit rude, as I didn't mean to, but I was just trying to prove my point.

Though I'm not sure if you are a different guest pretending to be the original answerer...

:)

Hello ilorty.  I saw that it was a quadratic, thus would have two roots.  For it to have "one" solution, those two roots would have to be the same, so that makes it something squared.  Then, it was just a matter of factoring an easy quadratic.  There had to be a 2 squared to obtain the 4 at the end.  I recognized that the factors would have to contain 2x to obtain the 8x in the middle (2 • 2x • 2). It takes longer to write it than it did to solve it, LOL.  I did take Algebra 1 & 2 but I don't remember discriminants, but from your example I understand now how it works. 

_.

I mean, all questions are copyrighted, so technically, no questions should be posted, using your logic. (Which would lead to the failure of this site, which will eventually turn into Facebook when the web2.0calc apocalypse arrives....what a scary thought...)

Also:

Are you actually Richard Ruscyzk, or are you impostering him? Or are you just a big fan of him?

x/60 + y/40 = 2, 

5x/40 + 3y/50 =11.7, solve for x, y

x =1380/ 17 =~81 + 3/17 - km

y = 440/ 17 =~25 + 15/17 - km.

$5,760 / $480,000 =1.2% Property tax rate on the first house.

1.2/100 x $600,000 =$7,200 Property tax on the $600,000 house.

Actually, guest, though you are right, there is a way to solve this without seperatly finding each and every one of the numbers. We can use counting ;)

no, counting is not the type of counting you learned in preschool, it's premutations, the kind you use for combonations. 

We can think of the number as 3 giant spaces:

–––––     ––––––    ––––––

The number in those amount of spaces is the possible numbers for the input:

    5              2              1

–––––     ––––––    ––––––

Now, all we have to do is multiply them all out:

5 * 2 * 1 = 10 possible numbers

And guest, you are correct. The 10 possible numbers are:

534  634  734  834  934   

543  643  743  843  943

:)

(a) Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number.  The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers.  They form a total of n + 2 numbers.  There are C(2,2) = 1 ways to choose two numbers from the 2 group.  There are C(n,1) = n ways to choose one number from the one group.  This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers.  They form a total of n - 1 numbers.  There are C(3,2) = 2 ways to choose two numbers from the 2 group.  There are C(n - 1,1) = n - 1 ways to choose one number from the one group.  This gives us a second term of 2*(n - 1).  The pattern will continue until we reach n.  So the two sides are equal.

abcd = 47.

We see that both angle MON and angle LOM make up angle LON. So...how does that help us? Well, we can form an equation!
 

133 = 2x + 33 + 3x + 20

133 = 5x + 53

80 = 5x

16 = x

Plug that back into 2x + 33:

16 ( 2 ) + 33 = 32 + 33 = 55

So no, guest, there is no possible way MON is 14. We we know that x HAS to be positive, and 14 is less than 33.

:)

The notation "~" means similar. This can help with the problem.

First, we need to find the scale factor of the two triangles. This is 30 / 18 = 5 / 3 (to get from the smaller triangle to the bigger one)

Therefore, we can say that:

5 / 3(x + 1) = 2x - 1

5 / 3 x + 5 / 3 = 2x - 1

5 / 3 = 1 / 3 x - 1

8 / 3 = 1 / 3 x

x = 8

We can also prove this backwards (AKA check our work)

2(8) - 1 = 15

x + 1 = 9

5 / 3 (9) = 15

And, we can see that our answer is right! (more of "my" answer, but "our" just sounds less narcissistic than "my": unless you already solved the problem, and got the same answer, but I doubt that since you posted this question)

:)

Actually, yes, there is something flawed with that logic.

Let's take a step back:

Instead of comparing them two, we can just pretend there is a guy that keeps on walking half his distance to get to his house. Let's pretend that he is one mile away from his house. So, here is how much space is left from him to his house:

1/2, 1/4, 1/8, 1/16, 1/32

Sure, it's going to take him forever (literally forever, not just hyperbolicly), BUT, he WILL get there. That's just some number sense and calculus combined.

Now, I will describe you what is wrong with your logic.

Sure, the begining is right: Candice WILL be 100 miles away. Let's just run down the line: (the first is candice, and the second is liz)

After 4 hours: 100 : 0   

After 5 hours: 125 : 50

After 6 hours: 150 : 100

After 7 hours: 175 : 150

After 8 hours: 200 : 200

See? We have already landed on the same spot.

What you are thinking is that Liz keeps on cutting the distance in half, but nowhere in the problem did it say that. Just because there is a pattern in the begining, does not mean it will last. Sometimes, me included, mathmeticians get sidetracked based on false, groundless (not only groundless, but already proved wrong) ideas. 

We have now also gotten the answer: 4 hours

This can also be done algebraicly:

Let "h" be the amount of hours driven. 
 

100 + 25 h = 50 h

100 = 25h

h = 4

MISTAKE CORRECTED THANKS TO ABOVE GUEST! Congratulations 🎊🎉🎈🍾 

:)

Hmmm...right. @guest, this is a math question, and therefore it can be done mathematicaly. 

How to do without brute force: All you have to do is finish Algebra 1 (or Algebra 2, depending on how your class is taught). 

Using discriminants, we see that for there to be only one root, the discriminant has to equal 0. Zero, and ONLY 0. (wrote it two ways just in case)

The discriminant can be found with b^2 - 4ac. We know that:

b^2 = 64

4ac = 16a

So:

64 - 16a = 0

- 16a = -64

a = 4 

THE ANSWER ENDS HERE

@guest (the second one, not the first one that probably gave a random number): Now, may I ask, how did you right away see that (2x+2)(2x+2) was the answer? I'm not trying to be rude. Maybe you did not see (2x+2)(2x+2), but instead saw that when a = 4, it could be factored into 4(x + 1)^2. Maybe. Maybe you should create an account, (clairvoyant would be a nice user name) so that we could chat, and you could teach me how to see the answer right away 

:)

it's incorrect. 

534  543  634  643  734  743  834  843  934  943  Total =  10 such numbers

Find a non-zero value for a such that ax^2+8x+4=0 has only one solution. 

(2x + 2)(2x + 2) = 0   so   a = 4 

I can't explain how to derive this mathematically. 

When I looked at the problem, I just saw the answer. 

I guess you could call it one form of solving by brute force. 

_.

Angle MON = 14 degrees.