Calculus

I'm assuming the plane is in 3-space, but this easily generalizes...

P1 = (x1,y1,z1) P2 = (x2,y2,z2) P3 = (x3,y3,z3)

Let v1 = P2-P1 = (x2-x1,y2-y1,z2-z1), a vector from the origin and similarly let

v2 = P3-P2 = (x3-x2,y3-y2,z3-z2)

Then, for real constants s,t , the plane is spanned by D + s v1 + t v2 = 0, for some constant D

cot(x) = sqrt[cosec^2(x) - 1]

1:100 (cm:mm). 1:100 (m:cm)

!x-y!

first calculate x-y, then calculate av.

If x equals 2, then you first write out your equation 2x + 5 = y. You then fill out 2(2) + 5 = y. 2 x 2 = 4, so then 4 plus 5 equals 9.

5

(√2+1)x+(√2-1)x=6 with symbols

There is no "answer", X=6 is complete in itself. There is nothing to solve. What is the value of X? Well, it's equal to 6.

I think this means (x+7)squared + (y-5)squared =4squared, which represents a circle centred at -7,+5

123.5597022

∫ f(x)nf'(x) dx = f(x)n + 1/(n + 1) + C

n ≠ -1

C is the constant of integration.

11 million pairs of hiking boots were sold in 1992

x-y=x² only if y = x - x²

There are infinitely many pairs of values (x, y) which will satisfy 4x + 1.5y = 79

The equation can be rearranged:

4x + 1.5y = 79

⇒ 1.5y = 79 - 4x

⇒ y = (158 - 8x) ÷ 3

For any given value of x, plug it into the above equation and the corresponding value of y will be calculated. So, some pairs of values (x, y) that satisfy this are:

(-2, 58), (1, 50), (4, 42), (7, 34)

2X - 6X + 5 = 0

subtract 5 from each side

2X - 6X = - 5

do what it says on the left

- 4X = - 5

X = 5/4

check in original equation

2(5/4) - 6(5/4) + 5 = 0

10/4 - 30/4 + 5 = 0

- 20/4 + 5 = 0

- 5 + 5 = 0

0 = 0

checks

Yes because the exponents (powers) of the variables are 1.

I don't what you are asking, but both equations are parallel. They go by the same pattern (two y units up, 1 x unit right.) but one crosses the y axis at 0 and the other one crosses the y axis at negative 1.

Yes, they are.

If you solve 2x + 7y = 6 for y, you get y = (-2x + 6) / 7.

If you solve 7y + 2x + 6 = 0 for y, you get y = (-2x - 6) / 7.

As you can see, both equations have the same slope. Therefore they are parallel.

You can punch those equations into a graphing calculator and visually verify that they are indeed parallel.

For any quadratic equation a*x^2 + b*x + c = 0, the two solutions are: x = [-b ± sqrt(b^2 - 4*a*c)]/(2*a). The quadratic formula can always be used to generate the solutions to a quadratic equation, but there are other (sometimes simpler and faster) methods.

Factoring can be useful when there are 2 real rational solutions to the equation. Factoring will be easier and faster if you know your multiplication table.

Take for example the following: x^2 + 8*x + 15. What two numbers will multiply to get 15, and add to get 8: Well the factors of 15 are (1 and 15) or (3 and 5), and the 3 and 5 works, so it factors into (x + 3)*(x + 5) = 0. The solutions are x = -3 and x = -5. This works because if the (x+3) factor is zero, then the whole thing equals zero, and x = -3 makes the (x+3) factor equal zero. Same for the (x+5) factor.

Completing the square is another way: Take this one: x^2 + 6*x + 8 = 0. This one is not hard to factor, but use completing the square. A perfect square quadratic (x + a)^2 = x^2 + 2*a*x + a^2, so can we get it into this form. 2*a needs to equal 6, so a = 3 and a^2 = 9, so we have x^2 + 6*x + 9, so how do we get there from our equation: 9 = 8 + 1, so add 1 to both sides:

x^2 + 6*x + 9 = 1. Now we have a perfect square (x + 3)^2 = 1, and take the square root of both sides: x + 3 = ± 1, so x = -4, and x = -2.

Completing the square is actually the method used to come up with the quadratic formula.

56328694

0.075 each.

(X+9)(x-2)

Trying to use the quadratic formula on this problem is like trying to use a chainsaw to brush your teeth: painful, doesn't get the job done, and what the heck are you thinking?

Just add: 694+77+900=1671

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