Construct the line perpendicular to line NO at point P.
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OpenStudy (anonymous):
|dw:1368626872724:dw|
ganeshie8 (ganeshie8):
do you have compass and straight edge with you
OpenStudy (anonymous):
I've got the answer ... thank you anyways but I do have another question I could use some help with
ganeshie8 (ganeshie8):
glad to hear that:) sure ask
OpenStudy (anonymous):
which statement is true?
A. all rectangles are quadrilaterals.
B. all quadrilaterals are rectangles.
C. all quadrilaterals are parallelogram.
D. all quadrilaterals are squares.
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ganeshie8 (ganeshie8):
thats easy
ganeshie8 (ganeshie8):
A. all rectangles are quadrilaterals.
ganeshie8 (ganeshie8):
but not all quadrilaterals are rectangles. some may be squares, some may be kites.. etc... you get the point
OpenStudy (anonymous):
yes (: thank you
ganeshie8 (ganeshie8):
np
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OpenStudy (anonymous):
If EF = 9x+14, FG = 56, and EG = 250. find the value of x.|dw:1368628860077:dw|
OpenStudy (anonymous):
@ganeshie8
ganeshie8 (ganeshie8):
its something like this : E is your home, G is your school. F is a ice-cream shop in between
ganeshie8 (ganeshie8):
distance between E and F = 9x + 14
ganeshie8 (ganeshie8):
distance between F and G = 56
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ganeshie8 (ganeshie8):
total distance between E and G = 250
ganeshie8 (ganeshie8):
thats what is given. ok
OpenStudy (anonymous):
ok so how do I solve for x ?
ganeshie8 (ganeshie8):
EG = EF + FG
ganeshie8 (ganeshie8):
250 = 9x+14 + 56
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ganeshie8 (ganeshie8):
you can solve x now ?
OpenStudy (anonymous):
22 ?
ganeshie8 (ganeshie8):
wrong. try again
OpenStudy (anonymous):
20 ?
ganeshie8 (ganeshie8):
250 = 9x+14 + 56
250 = 9x + 70
180 = 9x
20 = x
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ganeshie8 (ganeshie8):
correct !
OpenStudy (anonymous):
If AOC = 49, BOC = 2x+10, and AOB = 4x-15, find the degree measures of BOC and AOB.
ganeshie8 (ganeshie8):
do u have a pic attached to this q
OpenStudy (anonymous):
|dw:1368629841184:dw|
OpenStudy (anonymous):
@primeralph
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