Calculus

Yes

The line x + y = 1 will be defined by the points that satisfy that equation. The x-axis travels horizontally, the y-axis travels vertically.

Step 1. Draw the axes on graph paper. Each line should be an integer.

Step 2. Plot points that satisfy the equation, like (0,1)(1,0)(2,-1)(1,-2)(3,-2)(-2,3)

Step 3. Draw a line through them with a pencil using a straight edge of some sort.

When one of the numbers is zero, that's where it crosses the other axis. You'll find that the line crosses the x-axis at 1 on the y-axis and crosses the y-axis at 1 on the x-axis.

no it doesn't.

3x2= 6

6-(4x-2)= 6--8= 6+8 = 14

14+7 = 21

1/5 + 1/5 + 1/10

need common denominator to add fractions; 10 looks good and 1/5 = 2/10

2/10 + 2/10 + 1/10

= 5/10

which is...........

= 1/2

The problem is incomplete, as b is a variable.

Ans: 1.04

Like what?

7 21/100 or 7.21

(2x + 8)/4 = (2x - 8)/3

Multiply both sides by 12 to get rid of the fractions: 3*(2x + 8) = 4*(2x - 8)

Simplify: 6x + 24 = 8x - 32

Subtract 6x from both sides: 24 = 2x - 32

Add 32 to both sides: 56 = 2x or 2x = 56

Divide both sides by 2: x = 28

You can do this by temporarily ignoring the fraction part of the mixed number as long as you put it back again when finished!

For example,

For 3 1/4 - 2:

Here we can simply do 3 - 2 which is 1, and then add the fraction back again = 1 + 1/4 = 1 1/4

12x - 28 = 7x - 63

Subtract 7x from each side:

5x - 28 = -63

Add 28 to each side:

5x = -35

Divide each side by 5:

x = -7

On your graph paper, after you draw and number the 'x' and 'y' axes, draw a straight line
that goes through the point [ y = 2 ] on the y-axis, and has a slope of [ 0.5 ).

Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c are all vectors.

First, multiply out the cross products. Since the cross product of two vectors is itself a vector, we'll give the cross products some names to make this a little easier to understand:

(bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d

(cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v

(axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u

=> (axb)n[(bxc)x(cxa)] = un[dxv]

(dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w

=> un[dxv] = unw = u1w1 + u2w2 + u3w3

Now replace u and w with their vector coordinates (notice that the negative sign is factored into the middle terms, so the variables are switched).

u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 + (a1b2-a2b1)w3

= (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1)

Now we need to expand the v terms back out:

(d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1- d3c3a1 + d3c1a3

(d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2- d1c1a2+ d1c2a1

(d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2

So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 + d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+ (a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2)

= d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2 a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2 a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2 a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1

Some of the terms cancel out, leaving us with;

= d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 + d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1 - d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3 a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1

Now factor out d1 , d2 , and d3

= d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)

Now we can factor out a dot product of ( d1 + d2 + d3):

= ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)]

(Remember, to keep from changing the value of the equation we still need to keep the terms grouped together so that they multiply by the correct d components.)

Now factor out all the "a" components within the brackets:

= ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1 a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2 a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3 a3{c2b1 - c1b2})]

= dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1- c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} + {c1b3- c3b1} + {c2b1- c1b2})]

And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} = (bxc), so we factor out (bxc):

= dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + ( a1 a3+ a2 a3 + a3 a3)]

= dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 + a3)]]

= dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)]

(from above, remember that d = (bxc) )

= (bxc)n(bxc)n a n a

= [an (bxc)]^2

The complex conjugate of a+bi is a-bi.

This is written as z* where z is a complex number.

ex.

z = a+bi

z* = a-bi

r = 3+12i

r* = 3-12i

s = 5-6i

s* = 5+6i

t = -3+7i = 7i-3

t* = -3-7i = -(3+7i)

What is the abs value of X plus 2 less than 9?

5.5 x 6 x 7 = 231

Convert the function into the general form y = mx + c, where 'm' is the slope.

2x + 4y = 8 : 4y = -2x + 8 : y = -x/2 + 2 :

The slope is therefore -1/2.

2

That expression is the square of (5x + 1), so both of its factors are the same.

No, Newton, an Englishman, and Leibniz, a German, were both accomplished mathematicians, who independently developed an early form of calculus.

18 = X + 5/2

multiply through by 2

36 = 2X + 5

subtract 5 from each side

31 = 2X

divide both sides integers by 2

31/2 = X

check in original equation

18 = 31/2 + 5/2

18 = 36/2

18 = 18

checks

If -x - 2 = 9, we can solve for x by first adding 2 to both sides:

-x - 2 + 2 = 9 + 2

-x = 11.

Next, multiply both sides by -1 to remove the negative and solve for x.

-(-x) = -11

x = -11.

Divide 2 and 4 by 2. The two turns into a 1, and the equation would then look like: 3x=1/2.

x=.166666666667