All Answers 358514 Answers

Very nice, Omi  !!!

Yes, that’s true.  Only a few have ever publicly recognized the thoughtful insights I’ve brought to the forum.  When this happens, it usually seems like they are stating the obvious. This is why I rarely self-aggrandize, because it seems like I’m stating the obvious.

2345? 

2345

Find QR-RD.

Note that  4  = 2^2

And 64  = 2^6

So...we have 

4^x  = 64

(2^2)^x  = 2^6

2^(2x)  = 2^6

Since the bases are the same, we can solve for the exponents

2x  =  6    divide both sides by 2

x  = 3

Note that by ASA, triangles ACN  and XCN  are congruent

This means that  AN = XN  =  5

And AN + XN  = AX   = 10

So...in triangle  AXB.....M  is the midpoint of AB  and N  is the midpoint of AX

But.....if these sides are split in a 1:1 ratio....then a segment  connecting MN  must be parallel to base  BX

Therefore....we have that

AN / MN  = AX / BX

5  / MN = 10 / 14

5 / MN  = 5 / 7   

This implies that  MN   = 7

OK. What is your question?

That's nice.  

Have a question?

Since QR  is parallel to ST.....then segment QR  is  parallel to  base ST

This sets up the following relationship :

PQ / QS    = PR / RT

8 / 6   = PR /RT

4 / 3  = PR /RT

So....we have 7  equal parts to PT

And  PR  is 4 of them

So   PR  =  (4/7) (10.5)   =   6

Your question asks for the least possible value for the angle A;

that suggests a particular value for x, not a range of values.

3 + 2x  < -3x - 7      add 3x to both sides, subtract 3 from  both sides

5x  < -10      divide both sides by 5

x < -2

Hi, I'm the person who originally asked the question. I have personally been stuck on this problem for a very long time, so I'm severely at a loss and cannot properly provide you extra information. I'm pretty certain that you do not need to find \(x\) and should instead focus on finding the least possible value of \(x\) which is \(m\) and the greatest possible value of \(x\) which is \(n\). Thanks and sorry if this doesn't help!

Never mind it is very easy to solve

Thank you, I will keep that in mind

Thank you CPhill and XSquaredFactor, here is ho I got my answer

Thank you everyone.. It helped a lot XSquaredFactor and CPhill, here is my thought process on this one after XSquaredFactor gave me a hint

Idk either 

I found that the centre of the circle is    \((a,\frac{9a-16}{3})\)      where a is int the set ot reals

and the radius sqared iis                   \(r^2= 10a^2-40a+\frac{370}{9}\)

so the equation of the circle is

\((x-a)^2+(y-\frac{9a-16}{3})^2= 10a^2-40a+\frac{370}{9}\)

I know that is right becasue I have graphed it but when i take it further I get in a mess.  :/

As follows:

.

No but you get mailed a popsicle for every answer you get right

You can do matrix calculations on the home page calculator here.  See the following example:

Why would we hurry when you cannot even be bothered presenting your problem properly?

\(a=900+0.1s \\ b=1200+0.15(s-8000)\\ \text{find s such that }b>a\\ 1200+0.15(s-8000) > 900 + 0.1 s\\ 1200 +0.15s - 1200 > 900 + 0.1s \\ 0.05s > 900 \\ s> 18000\)