it would be 16 times larger
        
             
        
        
        

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

Now the zero product property tells us that there are two cases where this is true,

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of 

, so 

 where 
![n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}](https://tex.z-dn.net/?f=n%5B%2Ftex%20%5Dis%20any%20integer.%5C%5CMeanwhile%2C%5C%5C%5Btex%5D10%5Csin%20x-3%3D0%5Cimplies%5Csin%20x%3D%5Cdfrac3%7B10%7D)
which occurs twice in the interval 

 for 

 and 

. More generally, if you think of 

 as a point on the unit circle, this occurs whenever 

 also completes a full revolution about the origin. This means for any integer 

, the general solution in this case would be 

 and 

.
 
        
             
        
        
        
Answer:
2/5m - 1/5
Step-by-step explanation:
2(1/5m - 2/5) + 3/5
= 2/5m - 4/5 + 3/5
= 2/5m - 1/5
 
        
                    
             
        
        
        
The pH of the weak acid is 3.21
Butyric acid is known as a weak acid, we need the concentration of [H+] formula of weak acid which is given by this equation :
![[H^{+}]=\sqrt{Ka . Ma}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D%5Csqrt%7BKa%20.%20Ma%7D)
where [H+] is the concentration of ion H+, Ka is the weak acid ionization constant, and Ma is the acid concentration.
Since we know the concentration of H+, the pH can be calculated by using
pH = -log[H+]
From question above, we know that :
Ma = 0.0250M
Ka = 1.5 x 10¯⁵
By using the equation, we can determine the concentration of [H+]
[H+] = √(Ka . Ma)
[H+] = √(1.5 x 10¯⁵ . 0.0250)
[H+] = 6.12 x 10¯⁴ M
Substituting the value of [H+] to get the pH
pH = -log[H+]
pH = -log(6.12 x 10¯⁴)
pH = 3.21
Hence, the pH of the weak acid c3h7cooh is 3.21
Find more on pH at: brainly.com/question/14466719
#SPJ4