All Answers 358091 Answers

The web2 scientific calc is located on the home page.  You could have used that. 

R+T = 75 T=14% R+14% = 75 R= 75-14% WHAT IS R

Other people are often better at interpreting questions than I am butI do not understand what you mean by 14%R+14%

14%R would mean 0.14*R but adding 14% does not make sense to me.  14% of what?

5 in a thousand 

If the other side is expressed as one fraction then you can just turn both sides upside down.

$$\\\frac{1}{v^2}=\frac{a}{b}\\\\
then\\\\
\frac{v^2}{1}=\frac{b}{a}\\\\$$

Since the question actually asks where the left side IS NOT EQUAL to 8

The answer is actually

All real x EXCEPT x=8

Have a look at this site.

16a²b - 24ab²

The value common to both terms is 8ab so 16a²b - 24ab² can be written as 8ab*(2b - 3a)

I'm guessing you want to factor this.

Look for two numbers that multiply together to make +9 and that added together make -10.

Factors of 9 are 1, 3, and 9.  The product of 1 and 9 is 9; the sum of 1 and 9 is 10.  However, we want the sum to be -10, so we need to have -1 and -9:

(x - 1)*(x - 9) = x² - 10x + 9

Expand the terms and simplify:

First term:   (x + 3)³ = x³ +9x² + 27x + 27

2nd term:    (3x - 1)² = 9x² - 6x + 1

Subtract the second from the first:   x³ + 33x + 26

This must equal: x³ + 4 so

 x³ + 33x + 26 = x³ + 4

33x + 26 = 4

33x = -22

x = -22/33 = -2/3

Check:

$${\left({\mathtt{\,-\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}\right)}^{{\mathtt{3}}}{\mathtt{\,-\,}}{\left({\mathtt{\,-\,}}{\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{2}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}^{{\mathtt{2}}} = {\frac{{\mathtt{100}}}{{\mathtt{27}}}} = {\mathtt{3.703\: \!703\: \!703\: \!703\: \!703\: \!7}}$$

$${{\mathtt{\,-\,}}\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}^{{\mathtt{3}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\frac{{\mathtt{100}}}{{\mathtt{27}}}} = {\mathtt{3.703\: \!703\: \!703\: \!703\: \!703\: \!7}}$$

Ok.

At constant speed the net force on the sled is zero (Newton's 1st law of motion).  The frictional force is given by μ*weight where μ is the coefficient of friction.

μ*52 = 36

μ = 36/52

μ ≈ 0.692

Let's see...first, cross-multiply...this gives us....

20 = hk

Notice that, since the left side is constant, if k decreases, h must increase to retain this constancy !!!

So....that's it......k and h move in "opposite" directions......if one increases, the other decreases.....and vice-versa  !!!

I assume we might have this:

(1/5)x+2 = 3 - (2/7)x     Add (2/7)x to both sides and subtract 2 from both sides

(1/5)x + (2/7)x = 1        Multiply both sides by 35 to clear the fractions

7x + 10x = 35                Combine like terms

17x = 35                        Divide both sides by 17

x = 35/17

Circumference = pi x Diameter

31.4 = 3.14 x D    divide both sides by 3.14

10 = D

Edited:  Thank you Chris, I can see where you get that from. I guess I did not give enough thought to the wording of the question.

multiply the coefficents = (12y+18)-(3y-15)

then add like terms = 9y+3=0 and seperate the y and subtract 3 to place it on the other side...

9y=-3 now divide by 9

y=-1/3

CPHill, thank you. Even though I forgot to write the correct function you understood and gave me a satisfying answer. Bless you.

x=8

(4x+8)/5 = 8        multiply both by 5, to cancel denomentor

(4x+8) = 40          not subract 8 from the left, then place on the right side

4x = 32                now just divide by 4

x = 8

I assume you possibly want to graph

y - 9 = (1/7)(x - 8)  =

y = (1/7)(x-8) + 9

3.5 is also 3 and 5/10 so it can be simplified to 3 1/2 because 5 is half of ten

The very first anonymous gave the BEST answer to this question.

I cut and pasted the exact question into the calculator.

It may not be the intended question but it is the question that was asked!

People should learn to use brackets properly and ask the question that they intend.

I think you want to evaluate this, if I'm not mistaken....

lim  (x^6 -1) / (x^3 + 1)

x → -1

Factor (x^6 -1) as (x^3 - 1) (x^3 + 1)   ...  so we have ....

lim  [(x^3 - 1)(x^3 + 1)] /((x^3 + 1)

x → -1

Simplify  by "cancelling" the (x^3 + 1) terms

lim  (x^3 - 1)

x → -1

Now, take the limit

[ (-1)^3 - 1] =  [-1 -1 ] = -2

$$\left(\dfrac{x}{2^4}\right)^{-5}\\\\\\
=\left(\dfrac{2^4}{x}\right)^{5}\\\\\\
=\dfrac{2^{20}}{x^5}\\\\$$