An equal sign   

write the chain of unit fractions for each 

5/8

7/8

4/12

6/12

8/12

4/6

2/6

Hi shadew,

Sorry I took so long to get back to you I have been busy today.

I am very glad that you asked - I like to get feed back :)

This is the long way I could have done     $$(x-6)^2$$

$$\\(x-6)^2\\
=(x-6)(x-6)\\
=x(x-6)\;\;-6(x+6)\\
=x^2-6x\;\;-6x-36\\
=x^2-12x+36$$

Now $$(x-6)^2$$     is a perfect square.   There is a short cut method.

$$\\(a\pm b)^2=a^2\;\pm\;(2*a*b)\;+\;b^2\\
so\\
(x-6)^2=x^2\;-\;(2*x*6)\;+\;6^2=x^2-12x+36$$

I sometimes use this method when I want to square numbers in my head

eg

$$\\23^2\\
=(20+3)^2\\
=20^2+(2*20*3)+3^2\\
=400+120+9\\
=529$$

3

x² + y² = 10x + 5y -25     subtract 10x, 5y from both sides, so we have

x^2 - 10x + y^2 - 5y  = -25     complete the square on x and y..... so we have

x^2 - 10x + 25  + y^2 - 5y + 25/4  = -25 +25 + 25/4     factor and simplify

(x - 5)^2 + ( y - 5/2)^2  = 25/4

So, this is a circle whose center is ( 5, 5/2) and whose radius is 5/2

Now if we wanted this in terms of y.....first subtract (x - 5)^2 from both sides

( y - 5/2)^2 = 25/4 - (x - 5)^2    take the positive and negative square root of both sides

y -5/2  = ±√[25/4 - (x - 5)^2]    add 5/2 to both sides

y = ±√[25/4 - (x - 5)^2] + 5/2

And that's everything in terms of 'y"

Yeah.....what you have done is pretty much the same as I have done.......sorry, unless there's something I don't see, that's all I've got !!!!

Thanks. I got the awnser like this:

$${\mathtt{x}} = {\mathtt{10}}$$

$${\mathtt{y}} = {\mathtt{3}}$$

$${f}{\left({\mathtt{a}}\right)} = {\mathtt{xa}}{\mathtt{\,\small\textbf+\,}}{y}{\left({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{a}}\right)}$$

$${f}{\left({\mathtt{6}}\right)} = {\mathtt{48}}$$

$${f}{\left({\mathtt{7}}\right)} = {\mathtt{61}}$$

The arc length - s - is given by

s = rΘ   where r is the radius and "theta" is in rads....so we have

s = 20cm(1/5 rad)  = 4 cm

Let x be the  number of questions answered correctly.....then, (10-x ) are the number answered incorrectly......and the questions answered correctly times their point value (10) plus the questions answered incorrectly times their point value (-3) must be greater or equal to 51

10x  + (-3)(10-x) ≥ 51   simplify

10x - 30 + 3x ≥ 51   combine like terms

13x  - 30 ≥ 51   add 30 to both sides

13x ≥ 81    divide both sides by 13

x ≥ about 6.2   → 7

Mmmm.....I get the same answer you did.....if we only answered 6 qusetions correctly that would be 6(10)- 3(4)  = 48 points.......and that's not enough !!

Maybe someome else sees something I'm missing???

Approximate value obtained as follows:

.

Replace the sqrt(11) term with sqrt(29) and you'll get the book's answers.  

Don't forget that sqrt(29) can be positive or it may be negative. Try both. 

I'm confused. How did you get the -12x?

Sometimes it is labelled "log", in which case the log to base 10 key will be labelled "log10"

a=2i-j+k ; b =3k-i-2j find (2a-b, 2a+b)

$$\small{\text{
$
\begin{array}{rcl}
2a-b &=& 5i-k\\
2a+b &=& 3i-4j+5k
\end{array}
$
}}$$

Multiply both sides by 4

Let's see how StarMan may have arrived at that answer.

2/3rds of 276  = 184

And he only needs to read 20 pages tomorrow to have read 2/3rds of the book.

So.....184 - 20 = 164.......and that's how many pages he read last week.

If the book is giving two ansers, most likely it is a quadratic formuloa, which either youre calculator can not do or you are inputting it incorrectly. If you can not solve the quadratice using your head, Use the formula:

4*2=8 Ans

This triangle is a "Pythagorean Triple"......i.e., 8 - 15 - 17

The tangent of angle P is  ±( 8 / 15), depending upon the quadrant.

4² = 16

√4 = 2

How many times bigger is 16 when compared to 2?

Same question:  what number times 2 will give you 16?

--->   8 times   --->  16 is 8 times bigger than 2.

(1/2)·√(2)/2  =  [1·√(2)] / [2·2]  =  √(2)/4

Distance Formula:     d  =  √[ (x2 - x1)² + (y2 - y1)² ]

(x1, y1)  =  (2, -1)  --->  x1 = 2, y1 = -1             (x2, y2)  =  (-4, 2)   --->   x2 = -4, y2 = 2

--->   d  =  √[ (-4 - 2)² + (2 - -1)² ]  

--->   d  =  √[ (-6)² + (3)² ] 

--->   d  =  √[ 36 + 9 ] 

--->   d  =  √[ 45 ]

--->   d =  √9·√5

--->   d  =  3√5

If the problem is  y  =  .5x - 2x - 16:

--->   .5x - 2x - 16  =  0

--->       -1.5x - 16  =  0                                 (simplify)

--->              -1.5x  =  16                                (add 16 to both sides)

--->                    x  =  -10 2/3                         (divide both sides by -1.5)

If the problem is  y  =  .5x² -2x - 16

--->   .5x² - 2x - 16  =  0

--->     x²  - 4x - 32  =  0                              (multiply both sides by 2)

--->   (x - 8)(x + 4)  =  0                              (factor)

--->  either  x = 8  or  x = -4