<h2>
Answer:</h2>

<h2>Step-by-step explanation:</h2>
<h2>Given :</h2>

<h2>To Find :</h2>
<h2>Solution :</h2>
We have to add 1 in numerator and -10 in denominator because
![\tt \frac{8}{101} , \frac{9}{91} , \frac{10}{81} , \frac{11}{71} ...[Given]](https://tex.z-dn.net/?f=%20%5Ctt%20%5Cfrac%7B8%7D%7B101%7D%20%2C%20%5Cfrac%7B9%7D%7B91%7D%20%2C%20%5Cfrac%7B10%7D%7B81%7D%20%2C%20%5Cfrac%7B11%7D%7B71%7D%20...%5BGiven%5D)

The difference is 1 in numerator so we add 1 and the difference is -10 in denominator so we subtract -10.
Answer:
A triangle has sides of lengths 9,7, and 12 is not a right triangle .
Answer:
x > -4
Step-by-step explanation:
Inequalities like these are solved the same way equations are solved, except when multiplying both sides you flip the less than/greater than sign:
3x + 7 > -x -9
3x + 7 - 7 > -x -9 -7 Subtract 7 from both sides
3x > -x -16
3x + x > -x -16 + x add x to both sides
4x > -16
>
divide both sides by 4
x > -4
In this case the inequality sign did not have to be flipped since we never multiply both sides by a negative number.
Answer:
Step-by-step explanation:
Since a log graph is with base 10 and a ln graph is with base e (2.something), the log x graph will clearly have smaller numbers (as, for example, log100=2 and ln100=around 4.6). In addition, you only have to multiply a number by e to increase the power by 1 but you have to multiply a number by 10 (which is significantly larger than e) to increase logx's power by 1, therefore proving that the log x graph will grow slower