The polygons are similar, but not necessarily drawn to scale. Find the vale of x. (Will post picture of the polygon in comments) Please help!!!
the sides of the similar figures will be in equal ratios
for example 16 to 12 will be the same as (x-1) to 6
Yeah. So its 6 ; 12 (x-1) ; 6 (y + 1) ; 21 32 ; 24
\[\frac{ 16 }{ 12 }=\frac{ x-1 }{ 6 }\]
Then I would do 16 (6) = x - 1 (12) right?
all those will be the same fraction, 32/24 = 1.3333 16/12 = 1.3333 (x-1)/6 = 1.3333 all the sides will be 1.333 times as a scale factor
so you can set any of the two fractions equal
right, you just had the grouping off a little 16*6 = 12*(x-1) distribute the 12 into the (x-1) first
so it would end up 96 = 12x - 12
correct
and from there it would be 108 = 12x then 9 = x Correct?
yep
what equation would you set up to find out what y is?
Would it be 32/24 = y+1/21?
yep, just when you are typing equations, group the numerator (y+1)/21, the way you typed it turns into \[y + \frac{ 1 }{ 21 }\]
Oh alright. Do you think you could check my work on a couple more? I just want to make sure I'm doing them correctly.
sure
ok here it is: What is the solution to the proportion? \[\frac{ 3y - 8 }{ 12 } = \frac{ y }{ 5 }\] I got a little confused on this one. I ended up with: 5 (3y-8) = 12y 15y-40 = 12y -40= -3y 13.33 = y
right, 40/3
or 13.33 if you want an approximate answer
so y is 40/3?
yes, you can always test and see by putting that in for y in the original equation, and check if both sides are the same number.
32/12 = 2.66666 40/15 = 2.6666 same
Alright alright. Only one more. Sorry to bother you. Given the proportion \[\frac{ a }{ b } = \frac{ 8 }{ 15 }\] what ratio completes the equivalent proportion \[\frac{ a }{ 8 }\]?
I ended up with a = 1.1
\[\frac{ a }{ b } = \frac{ 8 }{ 15 }\] 15a = 8b \[\frac{ a }{ 8 } = \frac{ b }{ 15 }\]
you cant solve for a value of a, there are 2 variables in 1 equation., you can solve like above. a/8 = b/15 a in terms of b
Oh ok I see. I thought you would replace a/b with a/8 so it would be set up like: a/8 = 8/15
i dont think so, just solve for a/8 from what was given
Alright. Thank you so much! I really appreciate the help!!
no problem, anytime
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