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Rainbow [258]
3 years ago
5

The ideal width of a certain conveyor belt for a manufacturing plant is 50 in. an actual conveyor belt can vary from the ideal b

y 7/32 in. find the acceptable widths for this conveyor belt. CHECK ALL THAT APPLY
a) x>=49 24/32
b)x<=50 7/32
c)49 25/32<=x<=50 7/32
d)x+7/32<=50
e)absolute value of x-50 <=7/32
Mathematics
1 answer:
Nastasia [14]3 years ago
8 0
50 + 7/32 = 50 7/32
50 - 7/32 = 49 25/32

answers are :
c.) 49 25/32 < = x < = 50 7/32
e.) |x - 50| < = 7/32
     

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<h2><DEF = 40</h2><h2><EBF = <EDF = 56</h2><h2><DCF = <DEF =40</h2><h2><CAB = 84</h2>

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In triangle DEF, we have:

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<EDF=56

<EFD=84

So, <DEF =180 - 56 - 84 =40 (sum of triangle angles is 180)

____________

DE is a midsegment of triangle ACB

( since CD=DA(given)=>D is midpoint of [CD]

and BE = EA => E midpoint of [BA] )

According to midsegment Theorem,

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and DE = CB/2 = FB =CF

___________

DEBF is a parm /parallelogram.

<u>Proof</u>: (DE) // (FB) ( (DE) // (CB))

AND DE = FB

Then, <EBF = <EDF = 56

___________

DEFC is parm.

<u>Proof</u>: (DE) // (CF) ((DE) // (CB))

And DE = CF

Therefore, <DCF = <DEF =40

___________

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HOPE \:  THIS \:  HELPS.. GOOD  \: LUCK!

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