All Answers 358087 Answers

64) (x^2/x-11)^1/3

we can simplify this by doing the following:

(x^2)^1/3 / (x-11)^1/3

x^(2/3) / (x-11)^(1/3).

66) (x^-2/3/y^-1/3)^15

Since we have negative exponents, we can change their positions as follows:

[y^(1/3)/x^(2/3)]^15

Now, we can further simplify by doing:

[y^(1/3)^15] / [x^(2/3)^15]

y^5 / x^30/3

y^5 / x^10

$${\frac{{{\mathtt{y}}}^{{\mathtt{5}}}}{{{\mathtt{x}}}^{{\mathtt{10}}}}}$$.

72) f(x)=3x, g(x)=x^2

f(g(x))...we basically input whatever g(x) is into any x in f(x)

f(x^2) = 3x^2.

g(f(x))...we basically input whatever f(x) is into any x in g(x)

g(3x) = (3x)^2 = 9x^2.

74) f(x)=3x^2+2, g(x)=2x

f(g(x)) = f(2x) = 3[(2x)^2]+2 = 3(4x^2)+2 = 12x^2 + 2.

g(f(x)) = g(3x^2+2) = 2(3x^2+2) = 6x^2+4.

f(x)=2x+5 & g(x)=x^2-3x+2

56) -3f(x)*g(x)

-3(2x+5)*(x^2-3x+2)

(-6x-15)*(x^2-3x+2)

-6x^3+18x^2-12x-15x^2+45x-30

-6x^3+3x^2+33x-30

There are no restrictions to the x values as seen above...therefore, the domain is all real numbers.

58) 5f(x)/g(x)

5(2x+5)/(x^2-3x+2)

10x+25/(x^2-3x+2)

Here, note that we have a polynomial in the denominator...therefore, we have to consider what values make the denominator equal to 0, which will ultimately produce something that is undefined...

(x^2-3x+2) = (x-2)(x-1)...set both of these equal to 0 -->

(x-2) = 0 & (x-1) = 0

x = 2 and 1...now we know that x cannot equal 2 and 1...so our domain is restricted to all real numbers except for x = 2 & x = 1.

thanx 

17 = 9 - 2y

Subtract 9 from both sides

8 = -2y    divide both sides by -2

-4 = y

3(x-2)=18x+4       simplifying the left side, we have

3x - 6  = 18x + 4   subtract 3x from both sides......subtract 4 from both sides

-10 = 15x              divide both sides by 12

-10/15  = -2/3 = x

The steps to what? To graphing? To solving for x?

$$\begin{array}{l}
\mbox{Depends on what n is.}\\
\mbox{If $n = 0$, then $n^\infty = 0$}\\
\mbox{If $n = 1$, then $n^\infty = 1$}\\
\mbox{If n equals anything else though, I believe the answer would be infinity.}
\end{array}$$

Yes!  Both are correct.

4.83% is a presentage

Now we can write it  ; $${\frac{{\mathtt{4.83}}}{{\mathtt{100}}}}$$

Then we can slove this problem simply

It is; $${\frac{{\mathtt{4.83}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{283\,000}}$$ =$${\mathtt{4.83}}{\mathtt{\,\times\,}}{\mathtt{2\,830}}$$ = $${\mathtt{13\,668.9}}$$

finaly answer is 13668.9

log (x+2)=log (3x-3)

(x+2)=(3x-3)

3x-3 = x+2

3x -x= 2 +3

2x    = 5

 x     =5/2

        =2.5

2.5 is the value to x

use that method we can solve problem surly

6-1*0+2/2

first we must solve the dividing part & multiply part

It is;

       6-(1*0)+(2/2)

    = 6-0+1

Then we solve that normally

It is;

      6-0+1

    =(6+1)-0

    =7-0

    =7

Finally answer is 7

16^(1/2)

=$${{\mathtt{16}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$

=$${\left({{\mathtt{4}}}^{{\mathtt{2}}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$

=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{1}}}{{\mathtt{2}}}}$$  =$${\mathtt{1}}$$

so that answer is 4

Start by doing the first part.

(4x10^-5)^{^2}(2.63x10¹⁷)÷6.7x10³ 

Raising something to the second power means to multiply it by itself

(4x10^-5)(4x10^-5)(2.63x10¹⁷)÷6.7x10³        

Now we can do 4 x 4 and 10^-5 x 10^-5, as long as we multiply those two results together at the end.

For example, say we have (10x5)(10x5) Here we can multiply 10x10, then 5x5, then multiply them together.

10x10=100

5x5=25

100x25=2500

We'll get the same result if we multiply in the parentheses first.

(10x5)(10x5)

(50)(50) 

2500

So, this shows that we can multiply 4 x 4 and 10^-5 x 10^-5 next.

(4x10^-5)(4x10^-5)(2.63x10¹⁷)÷6.7x10³  

(4x4 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³    

(16 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³  

Remember, it we're multiplying numbers with exponets and they have the same base, we can add the exponets.

(16 x 10^-5x10^-5)(2.63x10¹⁷)÷6.7x10³

(16 x 10^(-5)+(-5))(2.63x10¹⁷)÷6.7x10³

(16 x 10^-10)(2.63x10¹⁷)÷6.7x10³

Now we can do the same thing with the next set of parentheses.

(16 x 10^-10)(2.63x10¹⁷)÷6.7x10³

(16x2.63  x 10^-10x10¹⁷)÷6.7x10³

(16x2.63  x 10^(-10)+(17))÷6.7x10³

(16x2.63  x 10^(-10)+(17))÷6.7x10³

(42.08x10⁷)÷6.7x10³

Now we still have basically the same rule, but we subtract the exponets because we are dividing

(42.08x10⁷)÷6.7x10³

(42.08÷6.7 x 10⁷÷10³)

6.280597 x 10^(7)-(3)

6.280597 x 10⁴

So that's our final answer, or, we could multipy this out.

6.280597 x 10⁴

6.280597 x (10)(10)(10)(10)

6.280597 x (10000)

62805.97

p=k/t

$${\mathtt{p}} = {\frac{{\mathtt{k}}}{{\mathtt{t}}}}$$

k=pt

that is the answer

Do you mean solve this for k?

If this is what you mean, you just have to get "k" alone, or in other words, k = something.

p=k/t              ---Multiply both sides by t

pt = k

k = pt

For some reason I don't think ths is what you meant.

Reply back if it's not, supply a better question, and we will surely help you out. :)

The volume of the full bale is V = w*pi*d²/4  where w is width and d is diameter.

If damage depth is s then the undamaged inner cylinder has volume Vu = (w-2s)*pi*(d-2s)²/4 because the width reduces by s at both left and right, and the diameter reduces by s on "both" sides.

The damaged volume is therefore  the difference between these. Here, we have w = 4', d = 5' and s = (2/3)' (because 8" is 2/3 of a foot). So:

$${\mathtt{V}} = {\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{V}} = {\mathtt{78.539\: \!816\: \!339\: \!744\: \!831}}$$ cubic ft.

$${\mathtt{Vu}} = {\frac{\left({\mathtt{4}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{2}}}}{{\mathtt{4}}}} \Rightarrow {\mathtt{Vu}} = {\mathtt{28.157\: \!978\: \!598\: \!841\: \!850\: \!5}}$$ cubic ft.

Damaged volume Vd ≈ 78.54 - 28.16 = 50.38 cubic ft.

As a percentage Vd/V ≈ 100*50.38/78.54 = 64.15%

Thurs 7/8/14

What is n to the power of infinity

@@ End of Day Wrap :       Wed 6/8/14      Sydney, Australia     Time 7:50 pm        ♬

Hi all,

Today was deadly quiet.  A few good answers were given by Alan, Ramalka, AzizHusain and NinjaDevo.

Thank you.

Only a couple of posts that may have interest.

How to remember pemdas. 

There are 304.8 millimetres in a foot

$${\frac{{\mathtt{27\,800}}}{{\mathtt{304.8}}}} = {\mathtt{91.207\: \!349\: \!081\: \!364\: \!829\: \!4}}$$ 

There are 91.2 feet in 27800mm but neither of these conditions is square.

The square bit does not make sense.

Your question does not make sense.

How to remember it:

PEDMAS

Please Excuse My Dear Aunt Sally

or

(Ninja's favourite)

Parachute Expert My Dear Aunt Sally

$${\frac{{\mathtt{4.83}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}{\mathtt{28\,300}} = {\mathtt{1\,366.89}}$$

Use this method to solve a problem.....i mean first solve( ) then ÷ after× after -  then +....it makes you easier to solve ur poblem accurately...