Answer:
![\boxed{\sf \dfrac{41}{63}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20%5Cdfrac%7B41%7D%7B63%7D%7D)
Step-by-step explanation:
hi, it means that we have to compute
![\dfrac{2}{9}+\dfrac{3}{7}=\dfrac{7*2+3*9}{7*9}=\dfrac{14+27}{7*9}=\dfrac{41}{63}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B9%7D%2B%5Cdfrac%7B3%7D%7B7%7D%3D%5Cdfrac%7B7%2A2%2B3%2A9%7D%7B7%2A9%7D%3D%5Cdfrac%7B14%2B27%7D%7B7%2A9%7D%3D%5Cdfrac%7B41%7D%7B63%7D)
Answer:
480
Step-by-step explanation:
He have about 0 dime and 69 coins each
Answer:
Step-by-step explanation:
Given that there is a function of x,
![f(x) = 2sin x + 2cos x,0\leq x\leq 2\pi](https://tex.z-dn.net/?f=f%28x%29%20%3D%202sin%20x%20%2B%202cos%20x%2C0%5Cleq%20x%5Cleq%202%5Cpi)
Let us find first and second derivative for f(x)
![f'(x) = 2cosx -2sinx\\f"(x) = -2sinx-2cosx](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%202cosx%20-2sinx%5C%5Cf%22%28x%29%20%3D%20-2sinx-2cosx)
When f'(x) =0 we have tanx = 1 and hence
a) f'(x) >0 for I and III quadrant
Hence increasing in ![(0, \pi/2) U(\pi,3\pi/2)\\](https://tex.z-dn.net/?f=%280%2C%20%5Cpi%2F2%29%20U%28%5Cpi%2C3%5Cpi%2F2%29%5C%5C)
and decreasing in ![(\pi/2, \pi)U(3\pi/2,2\pi)](https://tex.z-dn.net/?f=%28%5Cpi%2F2%2C%20%5Cpi%29U%283%5Cpi%2F2%2C2%5Cpi%29)
![x=\frac{\pi}{4}, \frac{3\pi}{4}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B%5Cpi%7D%7B4%7D%2C%20%5Cfrac%7B3%5Cpi%7D%7B4%7D)
![f"(\pi/4)](https://tex.z-dn.net/?f=f%22%28%5Cpi%2F4%29%20%3C0%20and%20f%22%283%5Cpi%2F4%29%3E0)
Hence f has a maxima at x = pi/4 and minima at x = 3pi/4
b) Maximum value = ![2sin \pi/4+2cos \pi/4 =2\sqrt{2}](https://tex.z-dn.net/?f=2sin%20%5Cpi%2F4%2B2cos%20%5Cpi%2F4%20%3D2%5Csqrt%7B2%7D)
Minimum value = ![2sin 3\pi/4+2cos 3\pi/4 =-2\sqrt{2}](https://tex.z-dn.net/?f=2sin%203%5Cpi%2F4%2B2cos%203%5Cpi%2F4%20%3D-2%5Csqrt%7B2%7D)
c)
f"(x) =0 gives tanx =-1
![x= 3\pi/4, 7\pi/4](https://tex.z-dn.net/?f=x%3D%203%5Cpi%2F4%2C%207%5Cpi%2F4)
are points of inflection.
concave up in (3pi/4,7pi/4)
and concave down in (0,3pi/4)U(7pi/4,2pi)
I have no idea what to do in problem 1
but I will do problem 2
A. 2⁵ = 32 (2×2×2×2×2=32)
B. 4³/2 = 32 (4×4×4=64 and then 64÷2=32
the others are not equal to 32