All Answers 358268 Answers

1/12x240= 2

4m-9=2m-3 😃

N/A .........😁

It's just 240/12

Indeed 1/x of y is y/x for all y and all x ≠ 0

f(x) = -7x + 11, determine f(10)

f(10) = -7*10 + 11 = -70 +11 = -59

Here is a whole list of resources.

log₄3 = x   --->   4^x = 3   --->   (2²)^x = 3   --->   2^2x = 3   --->   √(2^2x) = √3  

                --->   (2^2x)^1/2 = (3)^1/2   --->   2^x = (3)^1/2    --->   log₂(3)^1/2  = x

Therefore,   log₄3 =  log₂(3)^1/2

1. 
   
   
   

   
   

   
   

   
   

   an equation between two variables that gives a straight line when plotted on a graph.

   
   

   
   

   
   

   
   
   
   

a linear equation is an equation that creates a straight line on a graph (y=mx+b) ~numbers

48 ~numbers

Since 2,500 * 8,000 is 20,000,000, you would move your decimal to the left and count the spaces to find the exponent.  

In order to convert it to scientific notation, you would move your understood decimal from the

20,000,000(.) < understood decimal to the left until you get 2.0.  

Then you would count the decimal spaces (7) and that becomes apart of your answer,

which is 2.0 x 10 ^ 7

YES anon understood what you wanted and gave you a correct answer!

I shall show you

$$\\\frac{2x-1}{x+2}=A+\frac{B}{x+2}\\\\
$multiply both sides by (x+2)$\\
2x-1=A(x+2)+B\\
2x-1=Ax+2A+B\\
2x-1=Ax+(2A+B)\\
$ equating co-efficients$\\
2=A\\
-1=2A+B\\
-1=4+B\\
B=-5\\
so\\
\frac{2x-1}{x+2}=2+\frac{-5}{x+2}\\\\$$

so maybe you owe anon an appology  

I think anon's logic should work but I would do it much more simply.

$$\frac{2x-1 }{ x+2}\\\\
=\frac{2(x+2)-1-4}{x+2}\\\\
=\frac{2(x+2)-5}{x+2}\\\\
=\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\
=2+\frac{-5}{x+2}$$

Don't give anon negative points for trying to help you!!!

There are 8 numbers after the point so the bottom of the fraction will be 1 followed by 8 zeros

$$\frac{92857142}{100000000}$$

$$\\2y+3\times4y+y=8\\
2y+12y+y=8\\
15y=8\\
y=\frac{8}{15}$$

Start by multiplying both sides by (x+2), giving 2x-1 = A(x+2)+B

Now, multiply out and equate x terms on the left to x terms on the right, and we have A=2, and also -1=2A+B

Now just solve for B. 🚲

Oh yes ~Bio~

20 1/3 first you have to divide 61 / 3 then you multiply 3 * 20 

greg's age = a

4a - 3 = 3a + 7

4a = 3a + 10

a = 10

Greg is ten years old.

(How much money they have) plus (how much more money they need) to get (the total cost of the trip).

They have $560.00.

Let's make  x  represent the amount of money they need.

Total cost is $1680.00.

(560) + (x) = (1680)

     560 + x = 1680

Subract 560 from both sides:

             x  =  $1120.00

It's reasonable because they need more than $1600.00 and they have less than $600.00, so they will need more than $1000.00.

what is 5!

5! = 5 * 4! 

4! = 4 * 3!

3! = 3 * 2!

2! = 2 * 1!

1! = 1 * 0!

0! = 1

we have: 5! = 5 * 4 * 3 * 2 * 1 * 1 = 5 * 4 * 3 * 2 * 1

greatest common divisors (GCD) of 23^31 , 23^17

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 23^{31} &=& 23^{31} &&&&&& \\
\text{Number 2:} & 23^{17} &=& 23^{\textcolor[rgb]{1,0,0}{17}} && && && \\
\hline \text{The lowest Exponent}\\ \text{ of each prime number} &\text{GCD}&=& 23^{\textcolor[rgb]{1,0,0}{17}}&&&&&&
\end{array}}}$$

That will equals to 14,400,000. I hoped that helped .

 3^13*5^17,2^12*7^21 Least Common Multiple (LCM) ?

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 3^{13}*5^{17} &=& 2^0 &*& 3^{\textcolor[rgb]{1,0,0}{13}} &*& 5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^0 \\\text{Number 2:} & 2^{12}*7^{21} &=& 2^{\textcolor[rgb]{1,0,0}{12}}&*& 3^0 &*& 5^0 &*& 7^{\textcolor[rgb]{1,0,0}{21}}\\ \hline \text{The greatest Exponent}\\ \text{ of each prime number} &\text{LCM}&=& 2^{\textcolor[rgb]{1,0,0}{12}}&*&3^{\textcolor[rgb]{1,0,0}{13}} &*&5^{\textcolor[rgb]{1,0,0}{17}} &*& 7^{\textcolor[rgb]{1,0,0}{21}}\end{array}}}$$

3^13*5^17,2^12*7^21 Greatest Common Devisor(GCD) ?

$$\small{\begin{array}{l|rclclclcl}\text{Number 1:} & 3^{13}*5^{17} &=& 2^{\textcolor[rgb]{1,0,0}{0}} &*& 3^{13} &*& 5^{17} &*& 7^{\textcolor[rgb]{1,0,0}{0}} \\\text{Number 2:} & 2^{12}*7^{21} &=& 2^{12}&*& 3^{\textcolor[rgb]{1,0,0}{0}} &*& 5^{\textcolor[rgb]{1,0,0}{0}} &*& 7^{21}\\
\hline \text{The lowest Exponent}\\ \text{ of each prime number} &\text{GCD}&=& 2^{\textcolor[rgb]{1,0,0}{0}}&*&3^{\textcolor[rgb]{1,0,0}{0}} &*&5^{\textcolor[rgb]{1,0,0}{0}} &*& 7^{\textcolor[rgb]{1,0,0}{0}} = 1*1*1*1=1
\end{array}}}$$

You know for many people the teenage years are the most confusing years.

Perhaps that is why those years have the most confusing names ???   Could this be the answer?