Your digital camera has a 512 megabyte memory card. You take pictures at two resolutions, a low resolution requiring 4 megabytes of memory per photo and a high resolution requiring 8 megabytes of memory per photo. Write an equation to model the amount of each kind of photo you can take.

One possible answer is 8H + 4L = 512

Explanation:

H and L are placeholders for positive whole numbers, or 0 could replace either variable.

1 high resolution photo takes up 8 mb of space, so H of them take up 8H megabytes of space.

1 low resolution photo takes up 4 mb of space, so L of them take up 4L megabytes.

Overall, the two types of photos take up 8H+4L megabytes, which is equal to 512 since that's the max capacity. That leads to the equation 8H+4L = 512

Extra info:

There are other ways to express the equation 8H+4L = 512. We could divide each part by 4 to get 2H+L = 128, but I think this equation loses its descriptive quality in a way. We no longer can see that each high resolution photo takes up 8 megabytes (instead we might mistakenly think only 2 megabytes are used per high resolution photo). A similar mistake may happen with the low resolution photos also.

The equation 2H+L = 128 can be rearranged into L = -2H+128 when solving for L.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2