A photo is 6 inches long. Kamala enlarged the photo so that it is twelve inches long. By keeping the same proportion of the photo, what would be the width of the photo if the enlarged photo is 8 inches wide? Two images of rectangles. The first is six inches long and ? wide. The second is twelve inches long and eight inches wide Does this relationship represent direct or inverse variation? (1 point) Using the appropriate formula, find k. (1 point) Show your work and calculate the width of the original photo. (2 points)
can you help me @dpasingh
Sure @312856MLP
well i still m=need help with the Question above
This a kind of direct variation
ok what about the second one
Let the widht of the first image be k inches therefore: length of first image length of second image ----------------- = --------------------- Width of first image Width of second image i.e. \[\huge \frac{6}{k}=\frac{12}{8} \] \[\huge 6 \times 8 = 12 \times k \] \[\huge k= \frac{48}{12}\] K= 4 Hence the width of the first image is 4 inches. @312856MLP
what
@dpasingh
i'm here
i dont under stand what u mean
What you want to say?
well i have to do this so that my math teacher understands it and i know see would not understand that
How you want to solve it?
well it has to be like to the point a little kid would understand it
Well, this is the question of ratio and is related to direct variation: So the ratios of length and width of both images must be proportion, hence 6 : k = 12: 8 i.e. \[\huge \frac{6}{k}= \frac{12}{8}\] Now solve it for k
@312856MLP
it might be easier to understand if you compare the original length with the enlarged length to get from 6 inches to 12 inches had to multiply by 2 the width has to change by the same factor direct variation let k be the original width then 2k = 8 solve for k, original width = 4 inches
okay so what about the last one
let k be the factor = 2 let w be the width of the original = 4
thank yall very much @triciaal and @dpasingh thanks alot for the help
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