keeping in mind that anything raised at the 0 power, is 1, with the sole exception of 0 itself.
![\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{(r^{-7}b^{-8})^0}{t^{-4}w}\implies \cfrac{1}{t^{-4}w}\implies \cfrac{1}{t^{-4}}\cdot \cfrac{1}{w}\implies t^4\cdot \cfrac{1}{w}\implies \cfrac{t^4}{w}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bnegative%20exponents%7D%0A%5C%5C%5C%5C%0Aa%5E%7B-n%7D%20%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5En%7D%0A%5Cqquad%20%5Cqquad%0A%5Ccfrac%7B1%7D%7Ba%5En%7D%5Cimplies%20a%5E%7B-n%7D%0A%5Cqquad%20%5Cqquad%20a%5En%5Cimplies%20%5Ccfrac%7B1%7D%7Ba%5E%7B-n%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Ccfrac%7B%28r%5E%7B-7%7Db%5E%7B-8%7D%29%5E0%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7Dw%7D%5Cimplies%20%5Ccfrac%7B1%7D%7Bt%5E%7B-4%7D%7D%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20t%5E4%5Ccdot%20%5Ccfrac%7B1%7D%7Bw%7D%5Cimplies%20%5Ccfrac%7Bt%5E4%7D%7Bw%7D%20)
steesdpkfmspdnvhksdvnsdvspkjjhj pumpkingg
Answer:
The angle inscribed in a semicircle is a right angle. The inscribed angle theorem states that the angle θ, inscribed in a circle is half the measure of the central angle of the circle. So, if the given is a semi-circle, then the inscribed angle is half of 180, therefore, 90 degrees and a right angle
To find out the answer we divide 0.09 by 10:

÷10=0.009.
Therefore, 0.09 is 10 times as much as 0.009.
SInce the 'growth factor' is 1.32, the 'growth rate' is .32 or 32%