Multiply length times width to find the area. 6.5x11= 71.5
There are 4 aces in a deck of 52 cards. the odds of drawing one is 1/13. the probability of doing it again is still 1/13. the odds of doing it twice, as described, is then 1/13 * 1/13, or (1/13)², or 1/169 option A
Answer:
The number 10 is the base
Using distributive property, we get:
3x-9=0
3x=9
x=3
So, x=3. Now that was easy, huh? :)
Answer:
![\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7B%5Csqrt%5B4%5D%7B16a%5E%7B-12%7D%7D%3D2a%5E%7B-3%7D%3D%5Cdfrac%7B2%7D%7Ba%5E3%7D%7D)
Step-by-step explanation:
![16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}](https://tex.z-dn.net/?f=16%3D2%5E4%5C%5C%5C%5Ca%5E%7B-12%7D%3Da%5E%7B%28-3%29%284%29%7D%3D%5Cleft%28a%5E%7B-3%7D%5Cright%29%5E4%5Cqquad%5Ctext%7Bused%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%5Csqrt%5B4%5D%7B16a%5E%7B-12%7D%7D%3D%5Cbigg%2816a%5E%7B-12%7D%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Bused%7D%5C%20a%5E%5Cfrac%7B1%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5C%5C%5C%5C%3D%5Cbigg%282%5E4%28a%5E%7B-3%7D%29%5E4%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28ab%29%5En%3Da%5Enb%5En%5C%5C%5C%5C%3D%5Cbigg%282%5E4%5Cbigg%29%5E%5Cfrac%7B1%7D%7B4%7D%5Cbigg%5B%28a%5E%7B-3%7D%29%5E4%5Cbigg%5D%5E%5Cfrac%7B1%7D%7B4%7D%5Cqquad%5Ctext%7Buse%7D%5C%20%28a%5En%29%5Em%3Da%5E%7Bnm%7D%5C%5C%5C%5C%3D2%5E%7B%284%29%28%5Cfrac%7B1%7D%7B4%7D%29%7D%28a%5E%7B-3%7D%29%5E%7B%284%29%28%5Cfrac%7B1%7D%7B4%7D%29%7D%3D2%5E1%28a%5E%7B-3%7D%29%5E1%3D2a%5E%7B-3%7D%5Cqquad%5Ctext%7Buse%7D%5C%20a%5E%7B-n%7D%3D%5Cdfrac%7B1%7D%7Ba%5En%7D)
