NCTM December Math Problems Calendar!!

[koolwhip40]

1. If the area of a triangle with base 2F and height F-2 is equal to F, compute F.

4. If (2+2i)^2=a+bi compute the value of a+b

12. Given abs(11x-4)<or equal to 99/2 find the result when the largest possible solution for x is subtracted from the smallest possible solution.

16. At which alue of x does this equation take on its minimum value? y=(x-1)(x-19)+7

20. Let K=1/(1/4). If x=K/(K-x), find x.

24. Compute x+y, given the linear system with T cannot = 1 and with the ordered pair (x,y) as solution:
x+2y=3...Tx+(T+1)y=T+2
28. If 720 can be written as the product of five consecutive integers, determine the middle integer.

29. Compute the number of integers from 1 to 10000 inclusive that are divisible by 4 or 5.

For these problems I don't want the answers, I am looking for a way to look at the problem differently to start them off. Any help is very appreciated!

 

[tkhunny]

"Help" would require you to do somehting. Show what you can.

 

[Denis]

Can we get an email from your teacher confirming this?

 

[koolwhip40]

"Help" would require you to do somehting. Show what you can.

I switched my question at the bottom up to better clarify what I was asking. If you can help me with starting these problems it would be beneficial to me.

 

[Deleted member 4993]

1. If the area of a triangle with base 2F and height F-2 is equal to F, compute F.

Area of a triangle = 1/2 * (base) * (height)

replace the unknowns by given functions of F and then solve for 'F".

Please show us what progress you can make with above hint - then more will come

4. If (2+2i)^2=a+bi compute the value of a+b

12. Given abs(11x-4)<or equal to 99/2 find the result when the largest possible solution for x is subtracted from the smallest possible solution.

16. At which alue of x does this equation take on its minimum value? y=(x-1)(x-19)+7

20. Let K=1/(1/4). If x=K/(K-x), find x.

24. Compute x+y, given the linear system with T cannot = 1 and with the ordered pair (x,y) as solution:
x+2y=3...Tx+(T+1)y=T+2
28. If 720 can be written as the product of five consecutive integers, determine the middle integer.

29. Compute the number of integers from 1 to 10000 inclusive that are divisible by 4 or 5.

For these problems I don't want the answers, I am looking for a way to look at the problem differently to start them off. Any help is very appreciated!

 

[tkhunny]

"Help" would require you to do somehting. Show what you can.

I switched my question at the bottom up to better clarify what I was asking. If you can help me with starting these problems it would be beneficial to me.

Fair enough.

4. Perform multiplication, combine like terms, simplify

 

[koolwhip40]

1. If the area of a triangle with base 2F and height F-2 is equal to F, compute F.

Area of a triangle = 1/2 * (base) * (height)

replace the unknowns by given functions of F and then solve for 'F".

Please show us what progress you can make with above hint - then more will come

Ok I did that and I got 5/2 for the answer.

 

[Mrspi]

1. If the area of a triangle with base 2F and height F-2 is equal to F, compute F.

Area of a triangle = 1/2 * (base) * (height)

replace the unknowns by given functions of F and then solve for 'F".

Please show us what progress you can make with above hint - then more will come

Ok I did that and I got 5/2 for the answer.

I'm sorry...that is not the correct answer. And because we can't SEE the work you did, we have no idea of where you might be making a mistake. Please repost, and show us how you "got 5/2 for the answer."

 

[koolwhip40]

2

 

[koolwhip40]

I'm sorry...that is not the correct answer. And because we can't SEE the work you did, we have no idea of where you might be making a mistake. Please repost, and show us how you "got 5/2 for the answer."

Ok I looked back at what I did and I found the error I think..this is what I did for my answer..
.5(2F)(F-2)=F
F(F-2)=F
The F's cancel out because F=F so I was left with F-2=0, I added 2 to each side and got 2 for my answer.

 

[koolwhip40]

4. If (2+2i)^2=a+bi compute the value of a+b

I got a hint for this one but am I doing this correctly...4+4i+4i-4=a+bi...8i=a+bi...a+b=8.

 

[Mrspi]

I'm sorry...that is not the correct answer. And because we can't SEE the work you did, we have no idea of where you might be making a mistake. Please repost, and show us how you "got 5/2 for the answer."

Ok I looked back at what I did and I found the error I think..this is what I did for my answer..
.5(2F)(F-2)=F
F(F-2)=F
The F's cancel out because F=F so I was left with F-2=0, I added 2 to each side and got 2 for my answer.

While it is true that F = F, the Fs don't really "cancel out."

If you do the multiplication on the left side, you'll have

F[sup:2bsugowj]2[/sup:2bsugowj] - 2F = F

Now, let's get 0 on the right side, by subtracting F from BOTH sides:

F[sup:2bsugowj]2[/sup:2bsugowj] - 2F - F = F - F

F[sup:2bsugowj]2[/sup:2bsugowj] - 3F = 0

Can you continue from here? (Hint: the correct answer is NOT 2)

 

[koolwhip40]

While it is true that F = F, the Fs don't really "cancel out."

If you do the multiplication on the left side, you'll have

F2 - 2F = F

Now, let's get 0 on the right side, by subtracting F from BOTH sides:

F2 - 2F - F = F - F

F2 - 3F = 0

F(F-3)=0, since F cannot= 0, you would add 3 and F=3?

 

[Deleted member 4993]

F(F-3)=0, since F cannot= 0, you would add 3 and F=3?

Correct - now off to the next problem.

Use

\(\displaystyle (x \ + \ y)^2 \ = \ x^2 \ + \ y^2 \ + \ 2xy\)

 

[tkhunny]

Really, you're just throwing darts. Please seek local help for this. You need some personal tutoring and some face-to-face interraction.

RBGTHGANH

 

[koolwhip40]

(2+2i)^2...(2+2i)(2+2i)...using foil and the knowledge of i^2=-1..I got 4+4i+4i-4..This leaves me with 8i=a+bi...If i divide by i it leaves me with 8=a+b.

 

[tkhunny]

This is why I suggested that you need face-to-face. Your "divide by i" was a very poor attempt that can be cleared up nicely with a little personal discussion. That effort will NOT solve the problem. This problem is testing whether or not you know one specific concept. Obviously, you don't know it. You're making a good attempt, but you don't know it. Let's see if I can SAY it and have you translate to the arithmetic.

With some lack of formality:
Two complex numbers are identical if and only if their real parts are equal and their imaginary parts are equal.

1) There is no "divide by i" in there.
2) Let's see what you get.