The probability of rolling a 6 would be 1/6 or 16.67%
THat would be B.
The product of 22 polynomials is a polynomial.
Answer:
Factor GCF: 3x(2x⁴ - 5x + 4)
Step-by-step explanation:
Simply take out the greatest variable first and then take out the GCF of the numbers to find out factored GCF.
Answer:
1) ![f'(t)=4t,\ g'(t)=3t^2+4](https://tex.z-dn.net/?f=f%27%28t%29%3D4t%2C%5C%20g%27%28t%29%3D3t%5E2%2B4)
2) ![p(t) =2t^5+8t^3](https://tex.z-dn.net/?f=p%28t%29%20%3D2t%5E5%2B8t%5E3)
![p'(t)=10t^4+24t^2](https://tex.z-dn.net/?f=p%27%28t%29%3D10t%5E4%2B24t%5E2)
3) False
4)![q(t) =\dfrac{1}{2}t+2t^{-1}](https://tex.z-dn.net/?f=q%28t%29%20%3D%5Cdfrac%7B1%7D%7B2%7Dt%2B2t%5E%7B-1%7D)
![q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}](https://tex.z-dn.net/?f=q%27%28t%29%3D%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B2%7D%7Bt%5E2%7D)
5) False
Step-by-step explanation:
Given that:
and ![g(t) = t^3 + 4t](https://tex.z-dn.net/?f=g%28t%29%20%3D%20t%5E3%20%2B%204t)
Formula:
![1. \dfrac{d}{dx}x^n=nx^{n-1}](https://tex.z-dn.net/?f=1.%20%5Cdfrac%7Bd%7D%7Bdx%7Dx%5En%3Dnx%5E%7Bn-1%7D)
![2. \dfrac{d}{dx}C.f(x)=C.f'(x)\ \{\text{C is a constant}\}](https://tex.z-dn.net/?f=2.%20%5Cdfrac%7Bd%7D%7Bdx%7DC.f%28x%29%3DC.f%27%28x%29%5C%20%5C%7B%5Ctext%7BC%20is%20a%20constant%7D%5C%7D)
1) Using above formula:
![f'(t)=2\times 2 t^{2-1}=4t](https://tex.z-dn.net/?f=f%27%28t%29%3D2%5Ctimes%202%20t%5E%7B2-1%7D%3D4t)
![g'(t)=3t^{3-1}+4\times 1 t^{1-1}=3t^2+4](https://tex.z-dn.net/?f=g%27%28t%29%3D3t%5E%7B3-1%7D%2B4%5Ctimes%201%20t%5E%7B1-1%7D%3D3t%5E2%2B4)
2) ![p(t) =2t^2(t^3+4t)](https://tex.z-dn.net/?f=p%28t%29%20%3D2t%5E2%28t%5E3%2B4t%29)
Rewriting the formula by distributing the
term:
![p(t) =2t^2.t^3+2t^2.4t=2t^5+8t^3](https://tex.z-dn.net/?f=p%28t%29%20%3D2t%5E2.t%5E3%2B2t%5E2.4t%3D2t%5E5%2B8t%5E3)
![p'(t) = 10t^4+24t^2](https://tex.z-dn.net/?f=p%27%28t%29%20%3D%2010t%5E4%2B24t%5E2)
3) By using answers of part (1):
![f'(t).g'(t)=12t^3+16t](https://tex.z-dn.net/?f=f%27%28t%29.g%27%28t%29%3D12t%5E3%2B16t)
![p'(t) = 10t^4+24t^2](https://tex.z-dn.net/?f=p%27%28t%29%20%3D%2010t%5E4%2B24t%5E2)
Therefore it is <em>False </em> that ![p'(t) = f'(t).g'(t)](https://tex.z-dn.net/?f=p%27%28t%29%20%3D%20f%27%28t%29.g%27%28t%29)
4) ![q(t)=\dfrac{t^3+4t}{2t^2}](https://tex.z-dn.net/?f=q%28t%29%3D%5Cdfrac%7Bt%5E3%2B4t%7D%7B2t%5E2%7D)
Writing by distributing:
![q(t)=\dfrac{t^3}{2t^2}+\dfrac{4t}{2t^2}\\\Rightarrow q(t) =\dfrac{t}{2}+\dfrac{2}{t}\\\Rightarrow q(t) =\dfrac{1}{2}t+2t^{-1}](https://tex.z-dn.net/?f=q%28t%29%3D%5Cdfrac%7Bt%5E3%7D%7B2t%5E2%7D%2B%5Cdfrac%7B4t%7D%7B2t%5E2%7D%5C%5C%5CRightarrow%20q%28t%29%20%3D%5Cdfrac%7Bt%7D%7B2%7D%2B%5Cdfrac%7B2%7D%7Bt%7D%5C%5C%5CRightarrow%20q%28t%29%20%3D%5Cdfrac%7B1%7D%7B2%7Dt%2B2t%5E%7B-1%7D)
Using the formula:
![q'(t)=\dfrac{1}{2}t^{1-1}+2\dfrac{-1}{t^2}\\\Rightarrow q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}](https://tex.z-dn.net/?f=q%27%28t%29%3D%5Cdfrac%7B1%7D%7B2%7Dt%5E%7B1-1%7D%2B2%5Cdfrac%7B-1%7D%7Bt%5E2%7D%5C%5C%5CRightarrow%20q%27%28t%29%3D%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B2%7D%7Bt%5E2%7D)
(5)By using answers in part (1):
![\dfrac{g'(t)}{f'(t)}=\dfrac{3t^2+4}{4t}=\dfrac{3}{4}t+\dfrac{1}t](https://tex.z-dn.net/?f=%5Cdfrac%7Bg%27%28t%29%7D%7Bf%27%28t%29%7D%3D%5Cdfrac%7B3t%5E2%2B4%7D%7B4t%7D%3D%5Cdfrac%7B3%7D%7B4%7Dt%2B%5Cdfrac%7B1%7Dt)
![q'(t)=\dfrac{1}{2}-\dfrac{2}{t^2}](https://tex.z-dn.net/?f=q%27%28t%29%3D%5Cdfrac%7B1%7D%7B2%7D-%5Cdfrac%7B2%7D%7Bt%5E2%7D)
Therefore, it is <em>False </em>that:
![q'(t)=\dfrac{g'(t)}{f'(t)}](https://tex.z-dn.net/?f=q%27%28t%29%3D%5Cdfrac%7Bg%27%28t%29%7D%7Bf%27%28t%29%7D)
Answer:
Hi! Your answer here is 2550 GB. To find this answer, we multiply 255 (the number of kb) by 10,000,000 (the number of files.) This gives us 2,550,000,000 KB. There are 1,000,000 kb in 1 GB, so we divide 2,550,000,000 by 1,000,000, which gives us 2550. 2550 GB of files can be stored.