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3x + 3y + 9z = 30

-4x + 3y + 5z = 7

7x + 4z = 23

4kx + 4z = 3

If k = 7/4. 

4(7/4)x + 4z = 3

7x + 4z = 3, but 7x + 4z = 23??

That is impossible so it doesn't have any solutions. 

=^._.^=

Attn:  Bginner

Please, what did I do wrong?

Question: Dr. Worm leaves his house at exactly 7:20 a.m. every morning.
When he averages 30 miles per hour, he arrives at his workplace ten minutes late.  
When he averages 40 miles per hour, he arrives at his workplace five minutes late.  
When he averages 60 miles per hour, he arrives fifteen minutes early.
What speed should Dr. Worm average to arrive at his workplace precisely on time?

I combined what you told me with what is in this youtube clip (the clip made a lot of sense to me)

I'm almost certain this is a parallelogram. I can't visualize a way to assume it's a parallelogram and then change one of the unlabeled sides of the quadrilateral without breaking some of the given parameters. (bisection, lengths = 7, etc.)

You're absolutely right!!!

In order for them to have the same amount of money, Josh MUST give $28 to Ken!!!

But, in the following equation, Ken will "give" $28 to Josh!?!? It's weird!!! 

2/5x + 28 = 3/4x - 28

x = 160

Josh had $120              3/4(160) = 120        120 - 28 = 92        2/5(120) = 48

Ken had  $64                2/5(160) = 64           64 + 28 = 92         3/4(64) = 48

The ground is at height y = 0, so set y to zero and solve the resulting equation for t using the quadratic formula.  You will get two answers, but only the positive value is meaningful here.

When n is divided by 10, the remainder is a.
When n is divided by 11, the remainder is b.
What is n modulo 110, in terms of a and b?

\(\begin{array}{|rcll|} \hline n &\equiv& {\color{red}a} \pmod{10} \quad \text{or} \quad n = a+10r,~ r\in\mathbb{Z} \\ n &\equiv& {\color{red}b} \pmod{11} \quad \text{or} \quad n = b+11s,~ s\in\mathbb{Z} \\ n &\equiv& {\color{red}x} \pmod{110} \quad |\quad 110 = 10*11\\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline &n &=& a+10r \quad | \quad *11 \\ &\mathbf{11n} &=& \mathbf{11a+10*11r} \qquad (1) \\\\ &n &=& b+11s \quad | \quad *10 \\ &\mathbf{10n} &=& \mathbf{10b+10*11s} \qquad (2) \\ \hline (1)-(2): & 11n - 10n &=& 11a+10*11r - 10b - 10*11s \\ & n &=& 11a - 10b +10*11(r-s) \quad | \quad \text{let } r-s = t\\ & n &=& 11a - 10b +10*11t \quad |\quad 110 = 10*11\\ & \mathbf{n} &=& \mathbf{11a - 10b +110t} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{n} &=& \mathbf{11a - 10b +110t} \qquad (3) \\ n &=& \underbrace{\color{red}11a-10b}_{=x} \pmod{110} \\ \hline \end{array}\)

\(n \pmod{110} \equiv 11a-10b \)

Ok I've got this.       

Ok, that makes sense!

Yeah, you were right, I am in the US!

I understand the diagram and I will try to understand the acceleration part.

But, Thank you so much for helping me!

And, I am so sorry for sounding rude!

You are learning it by physics formulas rather than by calculus.  That is ok.  I think that is more common in the US than here.

(I assume you are in the US, apologies if I am wrong)

Still, it would be good for you to look closely and understand my diagram.

Then you will understand where the initial vertical velocity formula comes from. If you understand it then you don't have to memorize it.

For me the less formulas I have to memorize, the better.

Given is a quadratic xpression which is greater than equal to 0, thus, D = b^2-4ac <0
(2(4k-1))^2 - 4(15k^2-2k-7)<0
k^2-6k+8<0
(k-2)(k-4)<0
Thus, 2 or, k=3
 

;)

This question has already been answered here, https://web2.0calc.com/questions/probability_71218.

Let a and b with a > b > 0 be real numbers satisfying \(a^3 + b^3 = a + b\).
Find \(\dfrac{(a + b)^2 - 1}{ab}\).

\(Formula: \boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\)

\(\begin{array}{|rcll|} \hline a^3+b^3 &=& (a+b)(a^2-ab+b^2) \quad | \quad a^3 + b^3 = a + b \\ (a+b) &=& (a+b)(a^2-ab+b^2) \\ a^2-ab+b^2 &=& \dfrac{a+b}{a+b} \\ a^2-ab+b^2 &=& 1 \\ a^2+b^2-ab &=& 1 \quad | \quad a^2+b^2=(a+b)^2-2ab \\ (a+b)^2-2ab-ab &=& 1 \\ (a+b)^2-3ab &=& 1 \\ (a+b)^2-1 &=& 3ab \quad | \quad :ab \\\\ \mathbf{ \dfrac{ (a+b)^2-1 }{ab} } &=& \mathbf{ 3 } \\ \hline \end{array}\)

Let x be the number of small boxes and y be the number of large boxes. Both x and y must be whole numbers. The total mass of candy must be 220 pounds.

7x + 32y = 220

Find the expression of x in terms of y.

7x = 220 - 32y

\(x = {220 - 32y \over 7}\frac{}{}\)     
 

The value of x can be solved using trial and error. The value of x must be a whole number. Start from the largest value of y that is possible by equating x to 0.

7(0) + 32y = 220

\(y = {220 \over 32}\frac{}{}=6.875 \rightarrow y = 6\)
 

The largest value of y is 6. Start the trial and error from y = 6,5,4....

   y = 6 -> x = 220 - 32(6)/7 = 28/7 = 4 (Whole Number!)

From the first guess, the number of large boxes is 6 and the number of small boxes is 4.

The number of small boxes is 4.

3.13

Sorry typo!!!

-x^2 + 8x - 4

Do you know the Chinese Remainder Theorem?   You can apply it directly to this problem.

Anything divided by 0 is undifined. This means that 0 divided by 0 is also undefined.

I hope this helps!!!

-x^2 + 8t -4?

We were thought to put this in a formula x = (vcos(theta))t and y = -16t^2+(vsin(theta))t+h so, that's probably why I didn't understand the acceleration part!

Thank you for the wishes!

ok, you have found your answers, it might be correct.  I haven't checked.

You understand none of it.

That is not your fault, I am not blaming you, it is just an observation.

Maybe you will understand more as the topic progresses.

I hope that you do and I wish you well.

For anyone who can answer this question, if you drag the blocked by moderator image to another tab, you can see the picture there!