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Mathematics 13 Online
OpenStudy (anonymous):

SAT

OpenStudy (anonymous):

OpenStudy (anonymous):

how?

OpenStudy (vishal_kothari):

it is too long to explain...

OpenStudy (anonymous):

Oh, you don't always have to draw it

OpenStudy (vishal_kothari):

421/2=89/2 89/2-(1/2+3+1/2+1/2+3+1/2) 89/2-(8) 69/2=341/2

OpenStudy (vishal_kothari):

it will be 13...

OpenStudy (vishal_kothari):

in that gap..the horizontal woud be one more than the vertical.. let n is the horizontal one.. 3n+1/2(n-1)=69/2 n=10 for vertical ,it will be 9, so for the entire fence..it will be 9+4=13

OpenStudy (anonymous):

Each period is 7/2 feet. To make the full 42.5 foot fence, it takes 12 periods, plus one more vertical for a total of 13.

OpenStudy (vishal_kothari):

yeah..

OpenStudy (anonymous):

yup

Directrix (directrix):

@Sakigirl Scan the problem and the options. It is about counting posts. Think of the first post and 3 feet of railing as one component. Look at image SakiGirl1. That is one component. How many of these components are replicated along the fence? Well, the fence ends in a post but a component does not. From the fence length subtract ½ to get 42. To find out how many components are in the remaining 42 feet of the fence, Divide 42 by 3.5 or 7/2. The quotient is 12. Twelve components each of which has one post. So, that’s 12 posts. But, the very last post has not been counted because it was not part of a component. Add it to the 12 to get 13 posts as the answer.

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