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Mathematics 14 Online
OpenStudy (julyahx1):

An architect plans to make a drawing of the room of a house. The segment LM represents the ceiling of the room. He wants to construct a line passing through Q and perpendicular to side LM to represent a wall of the room. He uses a straightedge and compass to complete some steps of the construction, as shown below:

OpenStudy (julyahx1):

OpenStudy (julyahx1):

Which of these is likely to be his next step in constructing the perpendicular line? Fix the compass at Q, and draw an arc below the line. Fix the compass at L, and draw an arc above the line. Without changing the width, and position of the compass, draw an arc between L and M. Without changing the width of the compass, place the compass at M, and draw an arc.

OpenStudy (julyahx1):

@Will.H please help me

OpenStudy (julyahx1):

i think it's either A or C but I'm not 100% sure

OpenStudy (will.h):

it can't be C as drawing an arc positioned between L and M won't make a perpendicular line!

OpenStudy (julyahx1):

so then a right?

OpenStudy (will.h):

yeah

OpenStudy (julyahx1):

yay thanks for confirming! (:

OpenStudy (julyahx1):

i have another one i need help/confirmation on if ur not busy

OpenStudy (will.h):

my pleasure :)

OpenStudy (will.h):

sure

OpenStudy (julyahx1):

theres 3 more only that i need the confirmation/help on

OpenStudy (will.h):

Oki

OpenStudy (julyahx1):

Look at the right triangle ABC below:

OpenStudy (julyahx1):

OpenStudy (julyahx1):

A student made the following chart to prove that AB2 + BC2 = AC2: Statement Justification 1. Triangle ABC is similar to triangle BDC 1. Angle ABC = Angle BCD and Angle BCA = Angle DBC 2. BC2 = AC x DC 2. BC ÷ DC = AC ÷ BC because triangle ABC is similar to triangle BDC 3. Triangle ABC is similar to triangle ABD 3. Angle ABC = Angle ADB and Angle BAC = Angle DAB 4. AB2 = AC x AD 4. AB ÷ AD = AC ÷ AB because triangle ABC is similar to triangle ADB 5. AB2 + BC2 = AC x AD + AC x DC = AC (AD + DC) 5. Adding Statement 1 and Statement 2 6. AB2 + BC2 = AC2 6. AD + DC = AC What is the flaw in the student's proof? Justification 4 should be "AB ÷ AD = AB ÷ AC because triangle ABC is similar to triangle ABD." Justification 1 should be "Angle ABC = Angle BDC and Angle BCA = Angle DCB." Justification 2 should be "BC ÷ DC = BC ÷ AC because triangle ABC is similar to triangle BDC." Justification 3 should be "Angle ABC = Angle BAD and Angle BAC = Angle ABD."

OpenStudy (julyahx1):

i say the answers C

OpenStudy (julyahx1):

but I'm nostrum if the answers c or maybe a?

OpenStudy (julyahx1):

not sure*

OpenStudy (will.h):

Justification 1 should be "Angle ABC = Angle BDC and Angle BCA = Angle DCB." because they both right angles

OpenStudy (julyahx1):

so the answers not C?

OpenStudy (julyahx1):

you think its b?

OpenStudy (will.h):

am certain

OpenStudy (julyahx1):

you're certain its B? because i need to pass this class

OpenStudy (will.h):

yup

OpenStudy (julyahx1):

okay so 2 more

OpenStudy (will.h):

shoot

OpenStudy (julyahx1):

Kite ABCD is reflected over the line y = x. What rule shows the input and output of the reflection, and what is the new coordinate of A'?

OpenStudy (julyahx1):

OpenStudy (julyahx1):

(x, y) → (y, x); A' is at (2, −7) (x, y) → (−x, y); A' is at (7, 2) (x, y) → (−x, −y); A' is at (7, −2) (x, y) → (y, −x); A' is at (2, 7)

OpenStudy (julyahx1):

i think C, yes?

OpenStudy (will.h):

(x,y) --> (y,x) that's the rule of reflection across y=x A(-7,2) --> A'(2,-7) so it's A

OpenStudy (julyahx1):

oh ok makes sense

OpenStudy (julyahx1):

Rectangle ABCD is reflected over the x-axis, followed by a reflection over the y-axis, and then rotated 180 degrees about the origin. What is the location of point A after the transformations are complete?

OpenStudy (julyahx1):

OpenStudy (julyahx1):

(−5, 1) (5, −1) (−5, −1) (5, 1)

OpenStudy (julyahx1):

I say D, yes?

OpenStudy (julyahx1):

A?

OpenStudy (welshfella):

I think the answer to the first one is C, because its the first step towards getting a perpendicular line . The next step would be to bisect the angle formed by drawing a line between Q and the point between L and M.

OpenStudy (will.h):

(-5,1) --> (x,-y) --> (-5,-1) --> (-x,y) --> (5,-1) --> (-x,-y) --> (-5,1) yes A

OpenStudy (julyahx1):

which question is C @welshfella ?

OpenStudy (welshfella):

the first one

OpenStudy (will.h):

@welshfella wouldn't that be irrelevent to constrct an arc in the middle of that line?

OpenStudy (julyahx1):

this one? Which of these is likely to be his next step in constructing the perpendicular line? Fix the compass at Q, and draw an arc below the line. Fix the compass at L, and draw an arc above the line. Without changing the width, and position of the compass, draw an arc between L and M. Without changing the width of the compass, place the compass at M, and draw an arc. ??

OpenStudy (will.h):

yup he says it's C

OpenStudy (julyahx1):

@welshfella do u think the others were right or wrong that me and Will decided on ?

OpenStudy (welshfella):

this is what im thinking of |dw:1481895272702:dw|

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