![\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}](https://tex.z-dn.net/?f=%5Cbf%20%5Csqrt%7Bn%7D%3C%20%5Csqrt%7B2n%2B5%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsquaring%20both%20sides%7D%7D%7Bn%3C%202n%2B5%7D%5Cimplies%200%5Cleqslant%202n%20-%20n%20%2B%205%20%5C%5C%5C%5C%5C%5C%200%20%3C%20n%2B5%5Cimplies%20%5Cboxed%7B-5%20%3C%20n%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B-5%5Cleqslant%20n%20%3C%202%7D%7B%5Cboxed%7B-5%7D%5Crule%5B0.35em%5D%7B10em%7D%7B0.25pt%7D0%5Crule%5B0.35em%5D%7B3em%7D%7B0.25pt%7D2%7D)
namely, -5, -4, -3, -2, -1, 0, 1. Excluding "2" because n < 2.
I’m struggling too I’m sorry
Answer:
He bought 6 sweaters
Step-by-step explanation:
235.94-80.00=155.94
155.94/25.99=6
There 7 blocks of hundreds which means each such block is equivalent to 100.
There are 5 blocks of tens, which means each such block is equivalent to 10.
There are 8 blocks of ones, which means each such block is equivalent to 1.
The total of these blocks will be = 7(100) + 5(10) + 8(10) = 758
We can make several two 3-digit numbers from these blocks. An example is listed below:
Example:
Using 3 hundred block, 2 tens blocks and 4 ones block to make one number and remaining blocks to make the other number. The remaining blocks will be 4 hundred blocks, 3 tens blocks and 4 ones blocks
The two numbers we will make in this case are:
1st number = 3(100) + 2(10) + 4(1) = 324
2nd number = 4(100) + 3(10) + 4(1) = 434
The sum of these two numbers is = 324 + 434 = 758
i.e. equal to the original sum of all blocks.
This way changing the number of blocks in each place value, different 3 digit numbers can be generated.
Answer: Choice B
12.5 < x < 18.9
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Explanation:
We have a triangle with these side lengths:
- a = 10
- b = 16
- c = x = unknown
Let's assume that b = 16 is the largest side of this triangle.
By the converse of the pythagorean theorem, we need 
to be true in order for an acute triangle to happen.
So,
Now let's consider the possibility that the missing side x is actually the longest side.
Using the same theorem as before, we would say,
We found that x > 12.5 and x < 18.9
This is the same as saying 12.5 < x and x < 18.9
Put together, they form the approximate answer of 12.5 < x < 18.9
