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Ok. x = 3 is a vertical line.  This has infinite slope, so is basically undefined.

(I repeat: a horizontal line has zero slope).

BOSEOK is correct; the equation is x=3.

Contrary to what Gh0sty stated, the slope is defined when a linear function happens to be horizontal. Let's figure out the slope of the graph provided by Gh0sty. I will use the points \((3,-2)\) and \((0,-2)\). Let's find the slope.

Of course, the formula for slope is the following:

\(m=\frac{y_2-y_1}{x_2-x_1}\)

The slope is 0. 0 is a valid number for the slope. If you type into Demos y=0x-2, the line will appear exactly as the picture above. 

Let's try calculating the slope of the line x=-3. Two arbitrary points on the line are \((-3,2)\) and \((-3,0)\)

What is the slope of this? Let's try using the formula again. Let's see what happens.

Of course, any number divided by 0 is undefined and therefore it has an undefined slope. 

Therefore, BOSEOK's answer is correct because it meets both criteria of

1) A line with an undefined slope

2) A line that passes through the point \((3,-2)\)

And therefore, \(x=3\) is correct. 

To put it simply, in linear equations, if an equation is y= (any real number), then it is horizontal. For example, y=2 is horizontal, and it has an undefined slope. x=3 is actually a vertical line. Its slope is 0.

Correct me if I'm wrong. 

But my answeris x=3.

General equation of a straight line:  y = mx + c

1. 0 = -1*(-1) + c. Rearrange to find c

2. -2 = 3*6 + c. Again, rearrange to find c.

The slope of a horizontal line is well defined- it is zero.

I think what is wanted is as follows:

General equation of a straight line is y =mx + c

Here we know that this line goes through point (3, -2) so we can say

-2 = m*3 + c

This means c = -2 - 3m.  Put this in the equation for the line:

y = mx - 2 - 3m

or

y = m(x - 3) - 2

The slope m is undefined.

How can I know that line is horizontal?

y=-2. 

Because the line is horizontal, the slope is undefined. 

Creating a formula:

Let a be the length of each edge of the pot (area of a cube = a³)

Let c be cost per cubic foot

Cost to fill = a³ * c

1.5³ = 3.375 cubic feet

1.5³ * 4 = $13.50 to fill the planter

If the discontinuity occurs at, say, x = p, then integrate from a to p-eps, and add to the integral from p+eps to b, where a < p, b > p and eps is very small. If possible, do the integration symbolically, then take the limits of the results as eps tends to zero.

Incidentally, your f(x) example is continuous at x = 1: plot it to see!

Any multiple of the x^4 term will result in a power of x greater than x^3, so you can reduce this to (-3x + 2)^3

The only term that will produce an x^3 is the (-3x) term, so the coefficient of the x^3 term is just (-3)^3 or -27.

.

Total marks of boys = 35*50 → 1750

Total marks of girls = 46*g_ave

Total marks of class = (35 + 46)*40 → 81*40 → 3240

So we must have:  1750 + 46*g_ave = 3240

 46*g_ave = 3240 - 1750 

g_ave = (3240 - 1750)/46 ≈ 32.4

.

Thanks

This is the closest answer I got.

Suppose x is the common factor of the ratio, so

green marbles = 2x

red marbles = 5x

Equation:

2(2x+4) = 5x-10

    4x+8 = 5x-10

    8+10 = 5x-4x

 18 = x

 red = 90

green = 36

\(x+\frac{5}{x}-1+x+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-1+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-0+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}-8x=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{9}{{x}^{2}}-1-\frac{9}{{x}^{2}}\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{0}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=0-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=-1+1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=0\)

\({x}^{2}(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1)=0\times{x}^{2}\)

\(-6{x}^{3}+\frac{13{x}^{2}}{x}-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\)

\(-6{x}^{3}+{x}^{2}+13x-9=0\)

Since you cannot factor any further, the only way I know of to finish solving this equation is by graphing.

15 goes into 19 once and you have 4 left over, so 1 and 4/19 is answer

how i do the second one is double the first number, is that over the second, no then tripple it.

so 21+21 is 42, still smaller, 42+21 is 63, that went over so 42 is the biggest whole number part and it went 2 times.

what is left over between 42 and 62, you can subtract if you can not do it in your head.

62-42=20, so the answer is it goes in 2 times with 20/21 left over, so the final answer is

2 and 20/21

http://mathworld.wolfram.com/CentralAngle.html

square root of 28 is just a matter of breaking it down

2*2*7= 28 you have two 2's or its like square root of 4=2

however you want to see it, so the 2 can come out and you have 7*2 square root of 7

14sqrt7 

do the same with the other, you know the 7 has to be left cause the answers all have sqrt7

what times 7 =63 well 9, sqrt 9=3 so the 3 comes out

5*3=15 so you have 15sqrt7

like terms now, so 14-15=-1

answer is -1sqrt7 or C

sqrt12 is sqrt 4*3 so sqrt 4 is 2, 2 comes out, 3 stays in.

5*2=10, 10sqrt3, the m^3 is m*m*m so you can say two of those m's allow it to come out

only one x so it can not come out so 10msqrt3mx or D

one way to get the sqrt out of the denomiator or bottom of the fraction is 

to multiply by 1 or in this case sqrt13/sqrt13. this leaves 13 on the bottom and sqrt13 on the top

D is the only answer that meets that

cube root of 27 is 3 cause 3^3=27

can not really simplify the others

3^2/3 is cube root of 9

(x^2)^2/3 is x^4/3 or cube root of x^4

so cube root of 9x^4, i assume A is correct but mistyped?

THANK YOU SO MUCH YOU JUST SAVED MY BUTT FROM FAILING MY CLASS GOD BLESS YOU YOU ARE AN AMAZING HUMAN BEING DON'T LET ANYONE CHANGE YOU THANK YOU SO MUCH!!!!!

3)

This one is relatively simple. You have to take each modulus x some constant + remainder = each other as follows:

(6 * a) + r =(20 * b) + 9 =(45 * c ) + 4=x. You simply try, by iteration, to find the values of a, b, and c.

If you look at (45*c) + 4 and try c= 1, then it becomes: 45*1+4=49. and if you look at 20b+9, you will see that b= 2, and it becomes 20*2 + 9 =49. Check and see if it balances 6*a +r. If you take 49 and divide by 6, you will get 8, because 6*8 =48. And so 49 mod 6 =1. Therefore, r can only =1.

Since the LCM of 6, 20, 45 =180, therefore x will be:

x =180C + 49, and C =0, 1, 2, 3, 4, 5.......etc. So:

x =180*0 + 49 =49

x =180*1 + 49 =229

x =180*2 +49 =409

x =180*3+49 =589..........and so on. All these values of x satisfy the congruences.

2)

I don't actually get this question because there are so many unknown variables.

You may have to try trial and error to see if you get anything meaningful: For example, try some number such as 775:

775 mod 10 = 5

775 mod 13 = 8

775 mod 130 =120. Now, 5 and 8 are factors of 120, because:

120/5 =24 and 120/8 =15. Therefore:

(24*5)  +  (15*8) =120 + 120 = 240 !!!!.  Does it make sense???. I haven't done many of these for very long time!. 

Can you please help with Problem 2? And I don't really understand problem 3. But thanks anyways!!!!!

Solve for x :
(x + 5)/(x - 1) + (x + 8)/(x + 1) = (8 x + 9)/(x^2 - 1)

Hint: | Look for a polynomial to multiply both sides by in order to clear fractions.
Multiply both sides by x^2 - 1:
(x + 1) (x + 5) + (x - 1) (x + 8) = 8 x + 9

Hint: | Write the quadratic polynomial on the left hand side in standard form.
Expand out terms of the left-hand side:
2 x^2 + 13 x - 3 = 8 x + 9

Hint: | Move everything to the left-hand side.
Subtract 8 x + 9 from both sides:
2 x^2 + 5 x - 12 = 0

Hint: | Factor the left hand side.
The left-hand side factors into a product with two terms:
(x + 4) (2 x - 3) = 0

Hint: | Find the roots of each term in the product separately.
Split into two equations:
x + 4 = 0 or 2 x - 3 = 0

Hint: | Look at the first equation: Solve for x.
Subtract 4 from both sides:
x = -4 or 2 x - 3 = 0

Hint: | Look at the second equation: Isolate terms with x to the left hand side.
Add 3 to both sides:
x = -4 or 2 x = 3

Hint: | Solve for x.
Divide both sides by 2:
x = -4                   or                     x = 3/2

Note

(a + b)^3   =  a^3  + 3a^2b + 3ab^2 + b^3

(a + b) ^3  = a^3 + 3ab (a + b) + b^3 

(a + b) ^3   -   3ab (a + b)  =  a^3  + b^3

(a + b)  [( a + b)^2 - 3ab ]  = a^3 + b^3

(a + b) ( a^2 + 2ab + b^2 - 3ab)  = a^3 + b^3

(a + b) ( a^2 - ab + b^2 )  = a^3 + b^3

I haven't seen a problem like this before, so...I'm "flying blind"....but....I believe that we will have a pryramidal frustum here......

Triangles QCP and MDN are similar.....so...

DN /DM   =  PC / CQ

2 / 4  = PC / 8     →   PC  = 4

And the volume of a pyramidal frustum is  :

V  = (1/3)h  ( B1  + B2  + √ [ B1 * B2]  0

Here h is the frusum height, and B1 and B 2 are the areas of the bases

So  the height  = 16......the area of the top base , B1, =  (1/2)PC * CQ sin (60°) = (1/2) (4)(8) √3/2 =

8√3 ....and the area of the bottom base, B2 =  = (1/2)(DN) (DM) sin (60°)  = (1/2)(2)(4) √3 / 2  =

2√3

So....the volume is

(1/3) (16) ( 8√3 + 2√3  + √ [ 8√3 * 2√3] )  =

(16/3) ( 10√3 + √48 ) =

(16/3) ( 10√3 + 4√3)  =

(16/3) (14√3)  =  224√3 / 3   =  224 / √3   units^2