Melody

I have answered on the original thread. That is where you need to go if you want to continue with more questions or answers on this post.

Hi Dot.Box :)

Tertre has given you a very good hint. Thanks Tertre :))

As Tertre said, in similar figures the legths of corresponding sides must be in the same ratio.

So

\(\frac{JK}{MN}=\frac{LJ}{OM}\\ \frac{14}{3.5}=\frac{x}{12}\\\)

Now multiply BOTH sides by 12 to get the value of x.

What answer do you get?

--------

With the second one the two simialar trianges are   LOP and LMP

so

\(\frac{LO}{LM}=\frac{LP}{LN}\)

now do the substitution and solve for x

If you need more help you can ask on THIS thread but make you you say what you have done.

You can certainly do some of it on your own :)

The reason Gavin did it his way is that most times this would not work for a table of values.

You would not be able to pick the exact right x value to put into the table. BUT this one has been chosen so that it is easy.

f(x)=x^2-6x+8  this is a parabola, maybe you are meant to know that

g(x)=x-2   this is a line and maybe you are meant to know that as well.

There will be at most two points of intersection between a line and a parabola, lets see if we can find them

f(x)=x^2-6x+8

g(x)=x-2

I can see that for x=2     f(2)=0 and g(2)=0 

so f(x)=g(x) when x=2           So the first solution is x=2

Now continue the tables and see if you can find another match :)

I have deleted the bit where you gave the final answer Tertre.

Gavin's initial answer was very good.

But Gavin I was surprised that you use k instead of m for the gradient.

To my knowledge m is (almost) universally used for gradient 

I have seen a different letter used only in relation to statistics.

Maybe some countries are different.......

SamJones   behave yourself!

If you put as much effort into your statistics as you do into cheating you might get an A

That depends on what back to back means.

Why dont your draw it and think about the answer for yourself?

Thanks Gavin,

I know you meant well Tertre but I would like to encourage answerers NOT to give full answers.

If a person gives a part answer, in order to help the asker learn (just as Gavin has done) it is very frustrating when someone comes in and finishes it over the top of them. It diminishes their value as an educator.

We want to help the students learn not just spoon feed them the answers.

If the asker needs more help they should ask for it. 

Interactive learning is greatly encouraged!!

cos(pi-x) +sin(x-pi/2) = 1

Solve for all answers between 0-2pi

\(cos(pi-x) +sin(x-pi/2) = 1\\ -cosx+sinxcos\pi/2-cosxsin\pi/2=1\\ -cosx+0-cosx=1\\ -2cosx=1\\ cosx=-0.5\\ \text{x is in the 2nd or third quad}\\ \text{First quad }\pi/3\\ x=\frac{2\pi}{3}\quad or \quad x=\frac{4\pi}{3} \)

You have not even said what you think the answer is. 

Perhaps your are reading the answer section of your book upside down?

I stand by the answer that I have provided.

but if someone can see a flaw in my logic I will certainly listen to their reasoning. 

In a triangle the difference of any 2 sides must be greater than the lenght of the third side. 

Otherwise the 3 sides will not be able to join together!

Try drawing triangles on a peice of paper and try to be comfortable that this is the truth.

So 

42-33= 9

the third side must be more than 9.

Since it has an integer length, it's smallest vaue must be 10.

The perimeter is the sum of all the sides

42+33+9  is going to be the smallest possible perimeter.