Melody

why do we study math?, we don't use it in our daily life, yes we need + ,-,*,/ but we don't need the long solving and the difficult questions that drive me crazy!!

How much of what you learn at school, or even out of school, do you NEED in your normal life?

Well you need very basic arithmetic, and you need basic language.  Enough to fill ot a form or read a label on a tin of food.

These days it is not unreasonable that you need a few basic computer skill to cope as a fully functioning adult.

You need to know how to bath and dress yourself.  Mmm that is about it.  A bright person can have these skills by about the age of 8 or 9.  Do you suggest that the school leaving age be 8 and no-one needs to know anything outside of the necessities to get by on?  No one needs to extend themselves with language or art or music or mathematics or anything else.  Is that what you think?

All subjects teach learning skills that are independant of the obvious content.  Mathematics teaches you reasoning  and logic skill that you will not get from most other subjects.  That is why you are subjected to it so intensively.

Of course many people do use mathematics above this very fundamental level.  If you build or even order a new kitchen, or curtanis, or a new car then some mathematical skills are necessary or desirable. If you walk into a buiding or drive on a bridge a great many mathematics skills have been utilized in creating if for you.

MOST of all, mathematics is taught to all students for a considerable length of time because it develops their ability ability to think and reason!

I just used GeoGebra :)

Hi Chris,

Scaled pic

Drawing these pics to scale is always a time consuming but pleasant challenge :)

Hi Mellie,

In trapezoid ABCD, BC || AD, angle ABD =105°, angle A=43°, and angle C=141°. Find angle CBD, in degrees.

ok, the first thing I did is tried to work out what the trapezium looked like. 

THEN the answer became obvious.

BUT

it also became ovvious that I did not need to know what the trapezium looked like in ordet to answer this question.

Draw ANY trapezium where BC||AD

Label <ABD as 105 degrees.   (It doesn't need to LOOK anythink like 105 degrees, mine looks like an acute angle)

Label <A as 43 degrees 

Label < C as 141 degrees

Now you can see that <CBD is an alternate angle on parallel lines to <BDA. So they are congruent.

< BDA+105+43=180     Angle sum of triangle ABD

<BDA = 32 degrees

Therefore <CBD = 32 degrees

Here is a diagram most definitely not to scale.   

I am curious Chris, when I started answering this question there was no indication that you were already answering.

I do not know what time that was.

When you started answering, did you see me?  I suppose I had not logged in so it doesn't mean much.  I do not think I have ever seen a note saying that a guest is answering.  Maybe guests cannot see that anyone is already answering either???

I would like to ask you this question again when I am logged in properly. I want to know how it works. :/

If I had known you were already answering I would have moved on to another question...........

The area of trapezoid ABCD is 60. One base is 4 units longer than the other, and the height of the trapezoid is 5. Find the length of the median of the trapezoid.

Let the parallel sides be  x and x+4

The area of a trapezium is  the average or the parallel sides multiplied by the height

\(Area=\frac{(x+x+4)}{2}*h\\~\\ Area=\frac{(2x+4)}{2}*5\\~\\ Area=\frac{2(x+2)}{2}*5\\~\\ Area=5(x+2)\\ 60=5(x+2)\\ 12=x+2\\ x=10 \)

The parallel sides are 10 and 14 units

Now the median length is the average of the paralel side lengths

Length of median = (10+14)/2 = 12 units

what is three thirds of a metre?

Well if you cut a ribbon that is 1 metre long into 3 peices then each bit will be 1/3 of a metre.

If you have 3 of those, how many metres of ribbon will you have?

The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants.

The function has a line of symmetry at x=-3.

hence   -b/(2a)=-3                     b=6a

I might try looking at it a different way

you think (-3,0) should be a possible point on the parabola.

g(-3)= a(-3)^2 +b*-3 -6a

g(-3)= 9a -3b -6a = 3a-3b = 3(a-b) = 3(a-6a) =3(-5a) = -15a <0    since a must be positive.

Therefore (-3,0) cannot be a point on the parabola.

Yes, it has to be concave up because "a"  (the leading coefficient is positive)

Just like CPhill said :)

I had this one stashed away to come back to.  Well I am back !!

\(4^{2x+1}-10*4^x+4=0\\ 4*4^{2x}-10*4^x+4=0\\~\\ Let\;m=4^x\\ 4*m^2-10*m+4=0\\ 2m^2-5m+2=0\\ 2m^2-4m-m+2=0\\ 2m(m-2)-1(m-2)=0\\ (2m-1)(m-2)=0\\ 2m-1=0\qquad or \qquad m-2=0\\ 2m=1\qquad or \qquad m=2\\ m=\frac{1}{2} \qquad or \qquad m=2\\~\\ \)

\(\frac{1}{2}=4^x \qquad or \qquad 2=4^x\\ 2^{-1}=2^{2x} \qquad or \qquad 2^1=2^{2x}\\ -1=2x \qquad or \qquad 1=2x\\ x=\frac{-1}{2} \qquad \quad or \qquad x=\frac{1}{2}\\\)

Here is the graph