Calculus

6

It would appear that the domain is so very limited that the function may not be seen!

4

I know I know!!!! Its my fat juicy coc that you gotta slurp to get the answer.

The question is underspecified. As it stands, y can have any value.

x-9 = 15

x = 15+9

x = 24

If the question is, What x 3 equals 15?

Then 5 x 3 = 15

If the question is, What is 15 x 3?

Then 15 x 3 = 45

There are an infinite number of solutions.

Blank1 = 1 and blank2 = 15

B1 = -1 and B2 = -15

B1 = 2 and B2 = 7.5

B1 = 3 and B2 = 5

B1 = 1.1 and B2 = 15/1.1 etc, etc.

y=3x+4

-3x+y=4

x=-1/3

3 + 5x = 38

-> 5x = 35

-> x = 7

45

The answer is 44.

-6 * -11 = 66

66 + -22 = 44

-x2 + 11x + 26 = -(x - 13)(x + 2)

= (13 - x)(x + 2)

= (x - 13)(-x - 2)

Integration by Parts is a special method of integration that is often useful when two functions.

y = x - 15. We know x = 9.4

y = 9.4 - 15

y = -5.6

You use dry mix to answer because it means D: Dependant R: Respective and sooo on y:y Axis X:x axis

9 x 1760 = 15840.

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Yes, since (1,2) satisfies

y = x + 1

y = (1) + 1

y = 2

and

2 = 2

Changing velocity and constant acceleration? Yes.

Changing velocity indicates constant acceleration dv/dt = a constant(k) when v=kt.

Then dv/dt= dkt/dt= k. the constant k can be positive , negative or zero.

The equation for vertical motion is y = v0t + .5at2.

y is vertical displacement

v0 is initial vertical velocity

a is acceleration (in meters, normal gravitational acceleration is about -9.8 m/s/s, assuming positive y is upward displacement and negative y is downward displacement)

100 + 50 - 200 - 220 = -270

(ex-e-x)/2=8

ex-e-x=16

ex = 16-e-x

ln(ex) = ln(16-e-x)

x = ln(16-e-x)

x = ln((16ex-1)/ex)

x = ln(16ex-1) - ln(ex)

x = ln(16ex-1) - x

2x = ln(16ex-1)

e2x = eln(16e^x - 1)

e2x = 16ex-1

e2x-16ex +1 = 0

Consider ex as a whole to be a dummy variable "u". (ex=u) The above can be rewritten as:

u2-16u+1=0

u2-16u=-1

Using completing the square, we can solve this by adding 64 to both sides of the equation (the square of one half of the single-variable coefficient -16):

u2-16u+64=63

From this, we get:

(u-8)2=63

u-8=(+/-)sqrt(63)

u=sqrt(63)+8, u= 8-sqrt(63)

Since earlier we used the substitution u=ex, we must now use the u-values to solve for x.

sqrt(63)+8 = ex

ln(sqrt(63)+8)= ln(ex)

x = ln(sqrt(63)+8) ~ 2.769

8-sqrt(63) = ex

ln(8-sqrt(63)) = ln(ex)

x = ln(8-sqrt(63)) ~ -2.769

So, in the end, x~2.769 and x~-2.769. When backsubbed back into the original problem, this doesn't exactly solve the equation. Using a graphing calculator, the solution to this equation can be found to be approximately x=2.776 by graphing y=ex-e-x and y=16 and using the calculator to find the intersection of the two curves. This is pretty dang close to our calculated value, and rounding issues might account for this difference. The calculator, however, suggests that x~-2.769 is not a valid solution. This makes sense, and in fact it isn't a valid solution if you look at the graphs. This is an extraneous answer.

2xy + 4x + 8y + 16

= 2[xy + 2x + 4y + 8]

= 2[x(y + 2) + 4(y + 2)]

= 2(y + 2)(x + 4)

upper right is number one

then upper left is two

lower left is three

lower right is four

think of drawing a C on the grid