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combine like terms. 6c+-2c-3+d, 4c-3+d

That was a mistake, but I wanted to see how you guys would have gotten there. Sorry again! :(

To solve the equation \(0.00000000013284=13284\times10^{-x}\), we must realize that we are trying to get the same base. First, convert \(0.00000000013284\) into a scientific form. 

To do this, move the decimal point of \(0.00000000013284\) such that the decimal point moves in front of the first nonzero digit of the number. The number of times one moves is what the 10 will be raised to. Knowing this, \(0.00000000013284=1.3284\times10^{-10}\). We have now changed the equation to the following:
 

\(1.3284\times10^{-10}=13284\times10^{-x}\)

Now, let's continue moving that decimal of 1.3284 to the right. We'll move it 4 decimal places. This means that \(1.3284\times10^{-10}=13284\times10^{-14}\). Oh look! we have the same base now. Let's solve for x:

And there you go! 

Thanks Chris,

I had not seen that calculator before.  :/

1  km  = 10,000 dm

So

1km^2  =

(1km) (1km) =

(10,000 dm) (10,000 dm)  =

100,000,000 dm^2

I'm assuming that we are solving for v in the equation \(v^2=\frac{25}{81}\). Let's do that!

a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are colored so that a red 2, yellow 3, and green 4 is a different roll from red 3, yellow 4, and green 2.

These numbers are small enough so I can just count them

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Yea you can think about this one by thinking of the logic that happened with my first one.

I might think on it later. :)

bmichelle9, are you a naturally rude idiot, or have you received special training?

Put the question back so others can learn from it.  

I'll do what I can ....!!!

I used this tool for (24) ....http://keisan.casio.com/exec/system/14059932254941

The data was almost perfectly modeled as a quadratic

Here's the image....note that they use the odd form  Cx^2 + Bx + A  .....!!!

(25)   The equation that seems to fit best  is (roughly) :

y =    -15.9023229x^2 + 110.484418x + 3.015677  

Where x is the time in seconds and y is the height at that time

(26)  at x = 0....the height is ≈ 3.02 ft.

The velocity at this time is given by the  coefficient on the "x"  term ≈ 110.48 feet per second  

(27)  Set  y  = 6 and solve for x...so we have

6 =   -15.9023229x^2 + 110.484418x + 3.015677   subtract 6 from both sides and rearrange as

 -15.9023229x^2 + 110.484418x - 3.015677  = 0

I'll let you put this into the quadratic formula to solve for yourself.....

A = -15.9023229 

B = 110.484418

C = - 3.015677

Using Wolframalpha to solve this..

.

http://www.wolframalpha.com/input/?i=-15.9023229x%5E2+%2B%C2%A0110.484418x+-%C2%A03.015677+%C2%A0%3D+0

The outfielder has about   6.92 seconds

Thanks Chris, much appreciated.

I still have to think about it.

Maybe it is still too early in the morning for me. I have to have some excuse.   

How to carry out the theory of exponents with roots...??

In general.......  ⁿ √[ a^m ]  =  a^(m/n) 

And    a^(m/n) * a^(r/s)  =  a ^{(m/n + r/s)}

So...for instance.....suppose we had

⁵ √ [4³ ]  *   ³ √ [ 4²]  ......we can write this in exponential form as

4 ^3/5  * 4 ^2/3  =     4^{(3/5 + 2/3)}   = 4 ^{( [ 9 + 10] / 15 )}  =  4 ^(19/15)

P.S. -  Note that ......   √a  in exponential form is understood to be  a^(1/2)

\(x\) = cubic yards of cement the first truck can hold at one time

\(y\) = cubic yards of cement the second truck can hold at one time

\(f\) = number of trips of the first truck

\(s\) = number of trips of the second truck

\(f+s\) = the combined number of trips between the first and second trucks

\(x+y\) = the combined cubic yards of cement that can be held between the first and second trucks

\(xf+sy\) = the combined cubic yards of cement delivered between the first and second trucks

f+s= total # of trips

x+y= amount of cement they bring when they each go one trip

xf + sy = total number of concrete they bring after all their trips

The triangle just to the left of the blue area is a 45-45-90 triangle....and the side opposite the 90° angle has a length of  2√2 units  = √8  units

And  the blue area is a parallelogram with equal sides of this length

And the acute angle in the parallelogram = 45°

So....its area  =

2 * (1/2) (√8)^2 sin (45°)  =

8 / √2   units^2   =  8 √2 / 2  units^2  = 4 √2  units^2

\(6(\frac{4}{3}(14\div\frac{1}{7}))\div\frac{13}{10}= 6(\frac{4}{3}(14\cdot7))\cdot\frac{10}{13}= 6(\frac{4}{3}(98))\cdot\frac{10}{13}= 6(\frac{392}{3}))\cdot\frac{10}{13}=784\cdot\frac{10}{13}=\frac{7840}{13}\)or \(603 \frac{1}{13}\)

-16+9

17-24

Basically any expression that equals -7.

What is the value of 3/4y + 3/7x when x = 2 and y = 4 ?

(3/4)(4) + (3/7)(2)  =  

12/4  + 6/7 =

3 + 6/7    =     27 / 7

what is the value of X that makes the given equation true?

4x - 16 = 6(3 + x)  simplify

4x - 16 = 18 + 6x    subtract 4x, 18 from each side

-34 = 2x     divide both sides by 2

-17  = x

(-4)³ = -64

-64 + (-3) = -67

-67*(-4) = 268

268 - (-8) = 276

276 + (-3) = 273

¿Cómo escribes para publicar tu pregunta? 

The side of each square must be of length 4 ( to give areas of 16). The length of HC is therefore 2, as is that of BH.

Imagine a line from H to D.  Triangle HCD has area (1/2)*2*4 = 4.  Similarly triangle HDE has area 4. Hence area of polygon HCDE is 8.

Area of polygon ABHED is therefore 16 - 8 = 8

Add this to the area of square DEFG to get a total area of 16 + 8 = 24 for polygon ABHFGD.

Start by setting: \(u=\sqrt{11-x^2}\\ .\\u-\frac{2}{u}=1\\\)

Rearrange:  \(u^2-u-2=0\\.\\(u-2)(u+1)=0\)

So u = 2 or u = -1.  However, the original square root must be positive, so only the first value of u is valid.

\(\sqrt{11-x^2}=2\\.\\11-x^2 = 4\\.\\x^2=7\\.\\x=\pm \sqrt7 \)

Math is a basic necessity. Envisioning a world without it is downright unfeasible. 

There would be no clock or calendar. Therefore, any event that occurs on a schedule is impossible; there is no way to keep track of the time that passed or time in the future. 

Currency would be nonexistent. It only has value if the people accept its worth. But without math, currency would be worthless and useless.

All buildings would collapse, as they are governed by strict mathematical rules.

Technology as a whole would vanish for eternity. 

The world without math couldn't exist.

4/6 + 3/8 + 7 = 193/24 or 8 & 1/24

3/4 - 5/6 - 2/3 = -3/4

5(6/7)² = 180/49

5/8([4/5]/[1/3]) = 3/2

(3/7 - 2/3) + (4/5) (2/3) = 31/105

(4/6) + (3/8) +  7  =

2/3 + 3/8 +  7 =  [ common denominator is 24 ]

16/24 + 9/24 + 7 =

25/24 + 7 =

1 + 1/24 + 7 =

8 + 1/24

3/4 - 5/6 - 2/3 =   [common denominator  is 12]

9/12 - 10/12 - 8/12  =

-1/12 - 8/12 =

-9/12 =

-3/4

5 (6/7)^2  =

5 (6/7)(6/7)  =

5 (36/49) =

180 / 49

(4/5   /   1/3 ) * (5/8)   invert the second fraction and multiply

(4/5 * 3/1 ) * (5/8)  =

(12/5) * (5/8) =     swap denominators

(12/8 * 5/5) =

(12/8 * 1)  =

12/8  =

3/2

(3/7 - 2/3) + (4/5) (2/3) =    common denominator between the first two fractions is 21

(9/21 - 14/21) + (8/15)  =

(-5/21) + (8/15)   commond denominator  = 7 * 3 * 5  = 105

(-25/105) + (56/105)  =

31/105

We wouldn't be able to build anything; know which item at the store is cheaper than the other; measure the right amount of ingredients for anything you cook; technology wouldn't exist; we wouldn't be able to learn Math, Physics, or Chemistry in school; banking sectors, engineers, and architects would be unemployed; math professors on this planet would be sitting at home, thinking of something else to do in life.

That's what the world would be like without math.