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[ x² / (yz⁵) ]^(1/3) · z  

To rationalize the denominator under the radical sign, multiply the numerator and the denominator by y²z:

[ x²·y²z / (yz⁵·y²z) ]^(1/3) · z  

=  [ x²·y²z / (y³z⁶) ]^(1/3) · z 

=  (x²·y²z)^1/3 / (yz²)  · z 

=  (x²·y²z)^1/3 / (yz)

-9=12+x

-21=x

Set these equations equal to each other:

Since y = y   --->   (2/3)x + 3  =  (1/3)x + 6

Multiply both sides by 3:  2x + 9  =  x + 18         (to get rid of fractions)

--->                                   x + 9  =  18

--->                                         x  =  9

When x = 9:  y = (1/3)x + 6  =  (1/3)(9) + 6  =  3 + 6  =  9

They intersect in the point  (9, 9)

6.8=c-2.2

9=c

Sorry, Alisson, I don't understand the problem.

3x + y + 4x + y + 1

Rearrange the terms:  3x + 4x + y + y + 1

Group the like terms:  (3x + 4x) + (y + y) + 1

Simplify the like terms:    7x + 2y + 1

You can't add unlike terms, so this is the answer.

If you want to simplify a fraction that has exponents (with the same base), you can subtract the exponents.

Examples:  3⁵/3²  =  3³     and     10⁹/10²  =  10⁷

which means that  3²/3²  =  3⁰     And since 3²/3²  =  9/9  =  1,  3⁰  =  1.

This will be true for any base (except 0) and any exponent. 

Another example:  5⁴/5⁴  =  5⁰     And since 5⁴/5⁴  =  625/625  =  1,  5⁰  =  1

So, if x ≠ 0,  x⁰  =  1.

The reason that  0⁰ is not necessarily 1 is, in a problem like this:  0²/0²  is  0/0, and division by zero is not allowed.

1.   4 x 3/4

2.   4/1 x 3/4

3.   4x3/1x4

4.   12/4

5.   3

There are many possible answers.

Here is one of them:

You might notice that if you multiply the input by 3 and then add one, you can get the output.

--->   2 x 3 + 1  =  7

--->   3 x 3 + 1  =  10

--->   input x 3 + 1  =  output

33 1/3 % is one-third:  (1/3) x (30)  =  10.

A proper fraction is smaller than one, so we have to consider:  1/100, 2/100, 3/100, 4/100, ..., 99/100.

To be proper, it must be in lowest term, so we can remove all of them that have even numerators, leaving these 50 fractions:   1/100, 3/100, 5/100, 7/100, ..., 99/100 

Now, let us remove all those whose numerators are a multiple of 5: 5/100, 15/100, 25/100, 35/100, 45/100, 55/100, 65/100 75/100, 85/100, and 95/100. 

After removing these 10 fractions, there will remain 40 fractions.  

What good manners Billy - we really appreciate that.    

  (I gave you thumbs up just to help get you started )

If you give CPhill thumbs up then you will give him 5 points.  He will like that too :)

it can't intercect

you should ask your teacher if she/he can bump you up to the next level of math for instince im in 8th grade but i have 10 grade math

0.0526

______

         1

there is nada

6+5i+x=8+5i

YES what about it?     LOL

$$|1-6| \equiv 2 (mod 3)$$

This is true because |1-6|=5

and 5 divided by 3 is  1 remainder 2

2 is the remainder.

Modular arithmetic is only concerned with the remainder.

So I could say that

|a-b|-3n=2        where     $$n\in J \ge 0$$

The equation y=-(x+6)2+13 is given in vertex form, y=a(x−h)2+k.

x-(x-5) x=5

5-(5-5)

5-0

5

What grade are you in 

OK, shaniab.....!!!

15(.75)^5.....first, we would evaluate (.75)^5  = 0.2373046875....now ....mutiply this by 15 and you should get "unstuck"

Distance / Time  = Rate

Can you solve this based on that relationship ???