Start by multiplying 1.2 and 7.5 which will give you 9.Then you have 9-8/4. so then you do 8 divided by 4 which is 2. so you have 9-2 and that equals 7.
15 is you answer sir. the multiples are 15,30,45,60,75<span>,90,105,120,135,150,165,180,195,210,225,240,255,270,285,300,315</span>
I think it’s C
I hope it’s right
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)
Pythagorean Theorem is
a ²+ b ² = c ²
a and b are the shorter sides, while c is the hypotenuse, or longest side. Let’s say the side we are looking for is “b”.
Let’s first rearrange the equation, then sub in.
a ²+ b ² = c ²
b ² = c ²- a ²
b = √ (c ²- a ²)
b = √(17 ²- 15 ²)
b = √(289-225)
b = √ 64
b = 8
