The focal length of the mirror is 15 cm
From the question given above, the following data were obtained:
Object distance (u) = 30 cm
Image distance (v) = 30 cm
<h3>Focal length (f) =? </h3>
The focal length of the mirror can be obtained as follow:

<h3>Invert </h3><h3>f = 15 cm</h3>
Therefore, the focal length of the mirror is 15 cm
Learn more: brainly.com/question/1392083
<span>85% ethanol | 25% ethanol | 50% ethanol
x | y | 20 gal
use x and y because you don;t know how much she needs.
0.85x | 0.25y | 20(0.5)
85% is 85/100 or 0.85, and you need that much of x, same goes for the 25% and 50% mixtures so now you can make up 2 equations
1) x + y = 20 2) 0.85x + 0.25y= 10 (you get 10 when you multiply 20 by 0.5) now you can solve for x or y using substitution.
first rewrite 1) in terms of x or y: x+ y= 20 ----> y= 20 - x now you can substitute 20- x for y in the second equation.. 0.85x + 0.25y= 10 0.85x + 0.25(20-x)= 10 distribute here..(0.25 * 20 and 0.25 * (-x) ) 0.85x + 5 - 0.25x = 10 combine like terms 0.6x +5 = 10 move the 5 over to the other side 0.6x= 10 -5 0.6x = 5 divide both sides by 0.6 x= 25/3 or 8.3 now you know the amount of x so you can substitue this back into the first equation to find y. 0.85x + 0.25y= 10 0.85(25/3) +0.25y= 10 85/12 + 0.25y= 10 0.25y = 10- 85/12 0.25y= 35/12 y= 35/3 or 11.6 you can check by putting these values into the euations: 1) x+ y= 20 25/3 + 35/3 =20 20= 20 good so far 2) 0.85x + 0.25y= 10 0.85(25/3) + 0.25(35/3)=10 10 = 10
so our values for x and y work
x= 25/3 and y= 35/3</span>
Answer : The final temperature of the mixture is 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of iron = 
= specific heat of water = 
= mass of iron = 39.9 g
= mass of water = 
= final temperature of mixture = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get


Therefore, the final temperature of the mixture is 
Capture all of the smoke and weight it. it will weigh exactly the same before and after you burn it but will just be CO2 and H2O gas.