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)
According to the data i feel that the sign between the two sets is union therefore -1&5DONT COMEin first set therefore according to me ans is c
Answer:
48 students!
Step-by-step explanation:
12 / 1/4 = 12 x 4/1 = 48
hope i helped :D
20 and 14
Explanation:
Let the larger number be
x
Then the smaller number will be
x
−
6
Three times the smaller number is
3
(
x
−
6
)
, which is equal to two more than twice the other number which is
2
x
+
2
So,
→
3
(
x
−
6
)
=
2
x
+
2
Use the distributive property
a
(
b
+
c
)
=
a
b
+
a
c
→
3
x
−
18
=
2
x
+
2
Add
18
both sides
→
3
x
=
2
x
+
20
Subtract
2
x
both sides
⇒
x
=
20
We know that the larger number is
20
So, the smaller number is
20
−
6
=
14