CPhill

What is the conic for 4x^2-8x+144y+36y^2+4?

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We can write this as:

4x^2-8x+144y+36y^2  = -4          Dividing through by 4, we have

x^2 - 2x + 9y^2 + 36y = -1          Completing the suare on x and y, we have

(x^2 - 2x +1) + 9(y^2 + 4y +4) = -1  + 1 +36    Factoring, we have

(x - 1)^2   +9(y + 2)^2   =   36     Dividing though by 36, we have

[(x - 1)^2] / 36   + [(y + 2)^2] / 4   =   1

So we have an ellipse centered at (1, -2)  with a = 6 (the semi-major axis) and b = 2 (the semi-minor axis).

Tim has $1.50.......

If (x-1)(x-4)>0, why is x

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Have a look at the graph of this.....

Note, that we have three intervals of interest in this quadratic inequality.....

Namely, everything < 1, everything between 1 and 4, and everything > 4.

Note that all values between 1 and 4, inclusive, make this inequality untrue. For example, if x = 2, we have

(2-1)(2-4) = (1)(-2) = -2, and that's not greater than 0. Note that 1 and 4 don't work, either. They make the equality = 0, but it has to be >0.

So, the solutions to this inequality are .....x < 1    and x > 4

The lenght of a rectangle is twice its breadth and the perimeter is 6/7 m what is the area of the rectangle?

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OK, so the perimeter is given by 2(W + L), but L = 2W, so this simplifies to 2(W + 2W) = 2(3W) = 6W

So we have...

(6/7) m = 6W     dividing everything by 6, we have

W = (1/7) m      and L is twice this = (2/7)m

And the area = L X W   =   (1/7)m X (2/7)m  =  (2/49) m²

what is 2(a-10)=6a

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I think there's a slight error, here.

We have

2a -20 = 6a    subtract 2a from both sides

-20 = 4a         divide by 4 on both sides

-5 = a      or      a = -5

That's only found on "apple" computers......

Seriously, I think you meant "caret" symbol (for use as an exponent??)

It should be under the "6" on your keyboard

If you're talking about the calculator, just hit the x^y key

7x/2+7y/3=28/3 x/4+y/3=5/6 How would I solve for elimination?

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So we have

7x/2+7y/3=28/3

x/4+y/3=5/6

Let's multiply the first equation through on both sides by the common denominator of 2 and 3, (6), and let's multiply the second equation on both sides by the common denominator between 4, 3 and 6, (12). This gives

21x + 14y = 56  (Equation A)

3x   + 4y =   10  (Equation B)

Notice how this gets rid of those nasty fractions !! Now, I know there is a unique solution to this - don't ask me how I know, I just do - so let's proceed thusly:

Let's "eliminate" x........We can do this by multiplying the second equation by -7  on both sides. This gives...

21x + 14y =  56

-21x - 28y = -70      ......Now. just add the two equations together....this gives

-14y = -14     ........so it appears that y =1

Now, to find x , just substitute "1" for y in any of the equations - hint, I might pick Equation B - and see what x is.  You should check these values in all the equations to make sure they "work."

Re: How do you draw a quadratic function without plugging in numbers and plotting points?

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If you're taking Algebra, you will probably learn how to do some of this. If you take Pre-Calculus or Calculus, you will learn many other techniques that will enable you to "see" what a particular quadratic might look like. Barring that, unless it's a "basic" graph, you might have to be stuck with plotting (some) points. This could become tedious.

Here's a site where you can look at the graph of any quadratic (or any other polynomial in one variable) you specify:

Where

tan-1(x)=7/11

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We're just looking for an angle whose tangent = 7/11. Go to the on-site calculator, hit the 2^nd button, and then the "atan" button......key in your "7/11" and hit enter.

Note that this angle could also lie in the 3^rd quadrant, depending on the situation.