Calculus

Without further information, "y" can have any value.

red

y = 8

9 + x > 6

NB Treat the inequality like an 'equals' sign. There are two exceptions; they are when using fractions or dividing by negative numbers.

9 + x > 6

Subtract '9' from both sides

x > 6 - 9

x > -3

Verification ;

When x = -4

9 + - 4 = 5 < 6 ( Not true

When x = -2

9 + - 2 = 7 > 6 (True)

YES!!!!

Sin(2x) = Sin(x+x')

Sin(x+x') = SinxCosx' + CosxSinx'

I have put a 'dash' on an 'x' only to show its position in the identity. Both x & x' carry the same value.

Hence

SinxCosx' + CosxSinx' =

Sinx Cos x + Sinx'Cosx =>

2SinxCosx

NO!!!

x^2 + 1 = x^2 + 1^2 . This does NOT factor!!!!!

Remember , two squared terms with a positive between them, does NOT factor. However, two squared terms with a negative between them does factor.

e.g.

x^2 - 1^2 = ( x + 1)(x - 1) Note the change of signs inside the brackets.

So taking the Pythagorean equation ; a^2 + b^2 = c^2,, this does NOT factor. However, algebraically rearranging we have a^2 = c^2 - b^2 does factor to

a^2 = (c - b)(c + b)

-10

'x' is a fixed but unknown amount. It is not a VARIABLE!!!!

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From the information given the circle centre is (-4,3).

Since no sign is given with '4' , it is assumed it is positive.

The '= 4' is radius squared (Pythagoras).

So the whole equation becomes

(x --4)^2 + (y - 3)^2= 4 = 2^2 =>

( x + 4)^2 + ( y - 3)^2 = 2^2 (Note the Pythagorean equation).

Multiply out

x^2 + 8x + 16 + y^2 - 6y + 9 = 4

Collect terms

x^2 + y^2 + 8x - 6y + 21 = 0

The equation of the given circle.

x/7 = 2

Multiply both sides by '7'

7x/7 = 7 x 2 = 14

On LHS cancel down '7'

Hence x = 14

x = 13/11 & x = 17/19

Hence

x - 13/11 = 0 & x - 17/19 = 0

We now form a quadratic equation from these two linear equations.

(x - 13/11)(x - 17/19) = 0

Using FOIL multiply out.

F(First ; x X x = x^2

O(Outside) ; x X -17/19 = -17x/19

I(Inside ; -13/11 X x = -13x/11

L(Last ; -13/11 X -17/19 = (+)221/209

Hence the equation becomes

x^2 - (17/19 + 13/11)x + 221/209

Add the fraction in 'x'

x^2 - 434x/209 + 221/209 = 0

This is the quadratic eq'n. However, we can remove the denominator of '209' by multiplying through;-

209x^2 - 434x + 221 = 0

The equation 2x - 3y = 6 is a linear equation and a linear equation is always has a straight line as a graph

you subtract x from both sides for both equations to get it in y= form. so...

x+y=5

x+y-x=5-x

y=5-x

x+y=6

x+y-x=6-x

y=6-x

thanks:D but i meant with absolute value signs it was supposed to look like this : |x+y|=5 and |x|+|y|=6 ... but when i typed it in answers.com changed it

It is a binomial algebraic expression.

Its value can be evaluated if x is known.

y=54 if x=6

so we can write

y=9(x)

so y=k(x)

clearly y is directly proportional to x.

The source document states:

(S) The process takes three hours to complete.

The new document states:

(S) The process takes three hours to complete.

Which concept was used to determine the derivative classification of the new document?

• Extension
• Revealed by
• Contained in
• Compilation

3/(x + 5) = 2/(x + 1)

Multiply both sides by ( x + 5) & (x + 1)

Hence

3(x + 1) = 2)x + 5)

Multiply out the brackets

3x + 3 = 2x + 10

Subtract '2x' from both sides

x + 3 = 10

Subtract '3' from both sides

x = 7

Yes, this works, with one;

1 plus 1 over 1 = 2

No, it does not work with numbers over 2.

2 plus 1 over 2 = 1.5

3 plus 1 over 3 = 1.33333333...

4 plus 1 over 4 = 1.25

5 plus 1 over 5 = 1.2

3x + x - 4 = 0 Add all the 'x'.

Hence

4x - 4 = 0

Add '4' to both sides

4x = 4

Divide both sides by '4'

x = 1

3x-2y-7=0

C. p ‹ 12

This equals 3x+3 also written as 3(x+1). If you mean a linear equation like 3x + 3 = 0, then the solution is x = -1

Im assuming you are trying to come up with a y intercept equation for a graph. If that assumption is correct

x = y^2 - 2

y^2 = x + 2

y = SQRT ( x + 2 )