Rime factorization of 2001:
By prime factorization of 2001 we follow 5 simple steps:
1. We write number 2001 above a 2-column table
2. We divide 2001 by the smallest possible prime factor
3. We write down on the left side of the table the prime factor and next number to factorize on the ride side
4. We continue to factor in this fashion (we deal with odd numbers by trying small prime factors)
5. We continue until we reach 1 on the ride side of the table
<span>2001<span>prime factorsnumber to factorize</span><span>3667</span><span>2329</span><span>291</span></span>
<span>Prime factorization of 2001 = 1×3×23×29= </span><span>1 × 3 × 23 × 29</span>
Answer:
C. 25h – 5
Step-by-step explanation:
5(6h - 1) - 5h = 30h - 5 - 5h = 25h - 5
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)
If an object has a density of less than 
g/ml, the object will float in water.
So, fourth option is correct
The object with density less than g/ml will float, and the object with a density more than g/ml will sink.
So, if an object has a density of less than g/ml, the object will float in water.
Therefore, fourth option is correct
Learn more about density here:
brainly.com/question/3049000?referrer=searchResults
If you use this equation then you say that the ground is h=0 and solve as a quadratic.
The quadratic formula is (-b±<span>√(b^2-4ac))/2a when an equation is in the form ax^2 + bx + c
So the equation you have been given would be -16t^2-15t-151 = 0
This equation has no real roots which leads me to believe it is incorrect.
This is probably where your difficulty is coming from, it's a mistake.
The equation is formed from S=ut+(1/2)at^2+(So) where (So) is the initial height and S is the height that you want to find.
In this case you want S = 0.
If the initial height is +151 and the initial velocity and acceleration are downwards (negative) and the initial velocity (u) is -15 and the initial acceleration is -32 then you get the equation S=-15t-16t^2+151
Solving this using the quadratic formula gives you t = 2.64 or t = -3.58
Obviously -3.58s can't be the answer so you're left with 2.64 seconds.
Hope this makes sense.
</span><span>
</span>
