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As of the first one, you could say that a person drove 130 in 2 hours, therefore if you want to find the speed at which the person drove (mph/miles per hour) you divide both numbers with the number of hours traveled. 

calculation:

130/2

 2/2      

That draws you to the conclusion that the person was driving at a speed of 65mph (miles per hour).

As of the second, you could say that you paid $48 for 4 lbs of flour for example. If you then want to know the price of 1 lbs of flour you would have to divide both numbers by the amount of flour you paid $48 for. 

calculation:

48/4

4/4

This tells you that 1 lbs of flour cost $12. 

I hope this was helpful, let me know if I misunderstood the question. :)

CPhill,

Okay, I think I understand the logic. This approach will ONLY work if and when we have three isosceles triangles forming a parallelogram..if the triangles were anything else, this would not have worked, so it tells me that we will ALWAYS have the angles at 72 and 36 deg's....am I right by making this assumption?

I just wanted to express my gratitude once again for helping me, if I ever learned something new in mathematics, it certainly was this!!..Thank you very much and have a blessed day!

Hi Melody,

looking at your response I realise just how little I really know about mathematics...the types of maths available...like GeoGebra..I have never heard of it, this was a clever approach. sadly it did not work out this time,but still, it's a far above my knowledge base!

216.38334717810858

calculated by doing this

sqrt(sqrt(15673)*374)

15 divided by 3 is 5, therefore you also divide 9 by 3, and you end up with 3.

CPhill,

I have gone through your explanation, and I have to admit, I have NOT a clue what you have done, or why you approached the problem the way you did. I have never in my life seen anything like this...What you have done here is absolutely mind-blowing!..I'm trying to follow your steps, but I fail to understand why you did certain things. Anyways, I have double checked my labels, and they are all 100% correct, I have just neglected to say that PQRS was indeed a parallelogram. Sorry about that. Thank you for this extremely dificult explanation....I will be sure to study this, and who knows, maybe one day I'll understand the logic behind this!..Have a great day!!..

2^12

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Goodness me!!!!...what kind of math is that???..mama mia!!..

Melody, you have really gone the extra mile in trying to help me here!..I do appreciate this soooo much!, But this is my fault actually, because if I had mentioned in the beginning that PQRS was a parallelolgram, I'm sure you would not have spent so much time with this...Please accept my apology.

PES is definitely therefore a straight line...This one really has me speechless!

Hi Melody,

oops sorry, yes, PQRS is a parallelogram..sorry about that...but it still does not help me?

I think that's a really clever way to get the square of 108 !  

It was multiplied by   2 / 2     

cot θ  =  -2

                                         By the definition of cotangent, this means....

\(\frac{\cos\theta}{\sin\theta}=-2 \)

                                         Multiply both sides of the equation by  sin θ .

\(\cos\theta=(-2)\sin\theta\)

The Pythagorean identity says....

(sin θ)² + (cos θ)²  =  1

                                                 Plug in   (-2) sin θ   for   cos θ  .

(sin θ)² + ( (-2) sin θ )²  =  1

                                                 Mulltiply out the exponents.

sin²θ + 4 sin²θ  =  1

                                                 Combine like terms.

5 sin²θ  =  1

                                                 Divide both sides by  5 .

sin²θ  =  1/5

                                                 Since sin is negative in Quad IV, Take the negative sqrt of both sides.

sin θ  =  \(-\sqrt{\frac15}\,=\,-\frac{\sqrt1}{\sqrt5}\,=\,-\frac{1}{\sqrt5}\,=\,-\frac{\sqrt5}{5}\)

*edit*

Also note that the only Quadrants that have an angle with a negative cotangent are II and IV, so it would be impossible for the angle to be in Quad I or III .

It was multiplied by 2.

\(-24+5x=7x-2(-3x+4)\)

Distribute -2 into (-3x+4).

\(-24+5x=7x+6x-8\)

Combine like terms and reorder to make it easier to read.

\(5x-24=13x-8\)

Add 24 to both sides.

\(5x=13x+16\)

Subtract 13x from both sides.

\(-8x=16\)

Divide both sides by -8.

\(x=-2\)

Thank you so much. 🤙🏻 

No they are definitely not all independent triangles!

Can you work out what the equation of the line is?  

\(\frac1R=\frac1{2.3}+\frac1x \\~\\ \frac1R-\frac1{2.3}=\frac1x \\~\\ x(\frac1R-\frac1{2.3})=1 \\~\\ x=\frac{1}{(\frac1R-\frac1{2.3})} \\~\\ x=\frac{1}{(\frac{(2.3)1}{(2.3)R}-\frac{1(R)}{2.3(R)})} \\~\\ x=\frac{1}{(\frac{2.3-R}{2.3R})} \\~\\ x=\frac{2.3R}{2.3-R} \qquad\text{when }R\neq0\,.\)

Nah. I dont know ya

how did you get it?

3 yellow

Your question appears to want one to evaluate the expression \(\sqrt{(108b)^4}\). Let me try to make it easier so that the calculator is not necessary:

The other guy's answer really confused me because he didn't know how to use Latex, so I worked most of it out as best I could then asked this.

Hey! This is the same equation that you get from the last question.

You can solve it by factoring like this....

-2x² + x + 21  =  0

Multiply the  -2  and  21  to get  -42  .

Now split the middle term with two numbers that add to  1  and multiply to  -42 .

-2x² - 6x + 7x + 21  =  0

                                             Factor a  -2x  out of the first two terms and a  7  out of the last two terms.

-2x(x + 3) + 7(x + 3)  =  0

                                             Factor  a  (x + 3)  out of both remaining terms.

(x + 3)(-2x + 7)  =  0