Answer:
(0,2)
Step-by-step explanation:
i used a graphing calculator hopefully its right
1)180= 18 tens(18*10=180)
2)1600=16 hundreds(16*100=1600)
3)6000=6 thousands(6*1000=6000)
4)2700=27 hundreds(27*100=2700)
Hope this helps:)
Answer:
It’s A and C
Step-by-step explanation:
I’m pretty sure 0-O
Since
The square root of 11 is somewhere between 3 and 4. In order to round it to the nearest tenth, we have to try all numbers between 3 and 4 with one decimal digit, and see which is closest to 11 when squared. We have
![\begin{array}{c|c}n&n^2\\3&9\\3.1&9.61\\3.2&10.24\\3.3&10.89\\3.4&11.56\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7Dn%26n%5E2%5C%5C3%269%5C%5C3.1%269.61%5C%5C3.2%2610.24%5C%5C3.3%2610.89%5C%5C3.4%2611.56%5Cend%7Barray%7D%5Cright%5D)
So, the square root of 11 is somewhere between 3.3 and 3.4.
I’m assuming what you’re asking here is how to *factor* this expression. For that, let’s rearrange the expression into a more familiar form:
-c^2-4c+21
From here, we’ll factor out a -1 so that we have:
-(c^2+4c-21)
Let’s focus on the quadratic expression inside the parentheses. To find our factors (c + x)(c + y), we’ll need to find two terms x and y that multiply together to make -21 and add together to make 4. It turns out that the numbers -3 and 7 work out perfectly for that purpose (-3 x 7 = -21 and 7 + (-3) = 4), so substituting them in for x and y, we have:
(c + (-3))(c + 7)
(c - 3)(c + 7)
And adding back on the negative from a few steps earlier:
-(c - 3)(c + 7)
