Amir wants to scale up a photo to create a poster for his room. The original picture is 6 inches in width and 4 inches in height
2 answers:
So 6 in.w x 4 inches in.h
what will it be 30 inches wide
what we are doing right now is putting height over width, then the same thing to the other height over the width
now we cross multiply
(4) (30) = (6)(x)
120 = 6 x
divide both sides by 6 to isolate the variable
20 = x
The height of the poster, that has the width of 30 inches, will be 20 inches
Hope this helps :)
what will it be 30 inches wide
what we are doing right now is putting height over width, then the same thing to the other height over the width
now we cross multiply
(4) (30) = (6)(x)
120 = 6 x
divide both sides by 6 to isolate the variable
20 = x
The height of the poster, that has the width of 30 inches, will be 20 inches
Hope this helps :)
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Answer:
72 feet from the shorter pole
Step-by-step explanation:
The anchor point that minimizes the total wire length is one that divides the distance between the poles in the same proportion as the pole heights. That is, the two created triangles will be similar.
The shorter pole height as a fraction of the total pole height is ...
18/(18+24) = 3/7
so the anchor distance from the shorter pole as a fraction of the total distance between poles will be the same:
d/168 = 3/7
d = 168·(3/7) = 72
The wire should be anchored 72 feet from the 18 ft pole.
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<em>Comment on the problem</em>
This is equivalent to asking, "where do I place a mirror on the ground so I can see the top of the other pole by looking in the mirror from the top of one pole?" Such a question is answered by reflecting one pole across the plane of the ground and drawing a straight line from its image location to the top of the other pole. Where the line intersects the plane of the ground is where the mirror (or anchor point) should be placed. The "similar triangle" description above is essentially the same approach.
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Alternatively, you can write an equation for the length (L) of the wire as a function of the location of the anchor point:
L = √(18²+x²) + √(24² +(168-x)²)
and then differentiate with respect to x and find the value that makes the derivative zero. That seems much more complicated and error-prone, but it gives the same answer.
