9514 1404 393
Explanation:
"Like" radicals can be added and subtracted in the same way any like terms can be combined. It can be helpful to simplify the radical as much as possible so that it can be seen whether the radicals are "like" or not.
<u>Examples</u>:
√2 +√3 . . . . cannot be combined
√2 +√8 = √2 +2√2 = 3√2 . . . . the simplified radicals can be combined
If <em>c</em> > 0, then <em>f(x</em> - <em>c)</em> is a shift of <em>f(x)</em> by <em>c</em> units to the right, and <em>f(x</em> + <em>c)</em> is a shift by <em>c</em> units to the left.
If <em>d</em> > 0, then <em>f(x)</em> - <em>d</em> is a shift by <em>d</em> units downward, and <em>f(x)</em> + <em>d</em> is a shift by <em>d</em> units upward.
Let <em>g(x)</em> = <em>x</em>. Then <em>f(x)</em> = <em>g(x</em> + <em>a)</em> - <em>b</em> = (<em>x</em> + <em>a</em>) - <em>b</em>. So to get <em>g(x)</em>, we translate <em>f(x)</em> to the left by <em>a</em> units, and down by <em>b</em> units.
Note that we can also interpret the translation as
• a shift upward of <em>a</em> - <em>b</em> units, since
(<em>x</em> + <em>a</em>) - <em>b</em> = <em>x</em> + (<em>a</em> - <em>b</em>)
• a shift <em>b</em> units to the right and <em>a</em> units upward, since
(<em>x</em> + <em>a</em>) - <em>b</em> = <em>x</em> + (<em>a</em> - <em>b</em>) = <em>x</em> + (- <em>b</em> + <em>a</em>) = (<em>x</em> - <em>b</em>) + <em>a</em>.
Hey!
This could be rewritten as:
2x + 1 + 5x
We can add like terms together:
7x + 1
Answer:
V=2c
Step-by-step explanation:
![\\[Looking at the relationship of the numbers in the table, we can see which ones will work right off the bat.]\\[Our first numbers are both zero, so we cant do anything significant there.]\\[Next row, however, we can start doing the process of elimination.]\\\left[\begin{array}{ccc}1&2\\2&4\\3&6\\4&8\\5&10\end{array}\right]\\ \\[After plugging in the formulas to see if they're true, the only one that works every time is V=2c.]\\[That is your answer.]](https://tex.z-dn.net/?f=%5C%5C%5BLooking%20at%20the%20relationship%20of%20the%20numbers%20in%20the%20table%2C%20we%20can%20see%20which%20ones%20will%20work%20right%20off%20the%20bat.%5D%5C%5C%5BOur%20first%20numbers%20are%20both%20zero%2C%20so%20we%20cant%20do%20anything%20significant%20there.%5D%5C%5C%5BNext%20row%2C%20however%2C%20we%20can%20start%20doing%20the%20process%20of%20elimination.%5D%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%5C%5C2%264%5C%5C3%266%5C%5C4%268%5C%5C5%2610%5Cend%7Barray%7D%5Cright%5D%5C%5C%20%5C%5C%5BAfter%20plugging%20in%20the%20formulas%20to%20see%20if%20they%27re%20true%2C%20the%20only%20one%20that%20works%20every%20time%20is%20V%3D2c.%5D%5C%5C%5BThat%20is%20your%20answer.%5D)
