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Mathematics 16 Online
hardlyhuman:

Vocaloid

hardlyhuman:

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Vocaloid:

good, that looks correct to me

hardlyhuman:

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Vocaloid:

this problem stinks b/c EFG isn't a triangle on the diagram but I will assume they meant EFD

hardlyhuman:

Okay

Vocaloid:

ok I'm going to say that JKL and SRQ are not congruent everything else looks alright

hardlyhuman:

Okay thank you, I'll post the next one

hardlyhuman:

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Vocaloid:

yikes I've never really learned how to do this in my own classes so I will try and find something on the internet

hardlyhuman:

Okay, thank you.

Vocaloid:

so I believe the order you have posted is correct

hardlyhuman:

Okay, awesome. I'll post the next one

hardlyhuman:

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Vocaloid:

yeah I got 18 too well done

hardlyhuman:

Thank you :)

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Vocaloid:

I don't think that's right, try re-applying the slope formula (y2-y1)/(x2-x1)

hardlyhuman:

Okay

hardlyhuman:

Perpendicular means 90 degrees right?

Vocaloid:

yes (just find the slope of the original line first, accounting for perpendicularity is easy)

hardlyhuman:

Okay I'll try it

hardlyhuman:

7/6 is the slope of the original line. Is that right?

Vocaloid:

good now the perpendicular slope, we just flip the num and denominator then stick a negative sign in front so -6/7 = your answer

hardlyhuman:

oh really? its that easy. thank you, I'll post the next one

hardlyhuman:

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Vocaloid:

good

hardlyhuman:

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Vocaloid:

yup

hardlyhuman:

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Vocaloid:

yup

hardlyhuman:

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Vocaloid:

to get from step 5 to 6 you subtract m<3 from both sides so it would be subtraction property of equality otherwise a.o.k

hardlyhuman:

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Vocaloid:

I believe "open up the compass to length JK" needs to be step 2, right after drawing M everything else looks good

hardlyhuman:

Okay thanks

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Vocaloid:

should be "keep the compass opening + place the compass on F...etc." first choice since the original arcs must intersect the new arcs https://www.mathsisfun.com/geometry/construct-linebisect.html

hardlyhuman:

Okay

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Vocaloid:

the compass needs to be placed on G and then opened up all the way to H so the sides are all equal (choice C) sorry if my explanations on these construction problems are bad, I try not to give direct answers but I've never really learned these before so it's hard for me to explain ;_;

hardlyhuman:

I totally get it, I don't understand most of this stuff. I don't know why but no matter how much I pay attention it just won't stick. I really appreciate all your help.

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Vocaloid:

shorter side is one third the length of the longer side if the longer side is 3a then the short side is just a

Vocaloid:

|dw:1515807028537:dw|

Vocaloid:

so just (0,a) = your answer

hardlyhuman:

Okay, thank you.

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Vocaloid:

yup good

hardlyhuman:

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Vocaloid:

they're congruent, and A'B'C'D' compared to ABCD is moved to the left 7 units making answer 2 the only possibility

Vocaloid:

**** sorry answer 3

hardlyhuman:

thank you, would you mind helping me with a 3 question worksheet real quick please?

Vocaloid:

sure

hardlyhuman:

thank you

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Vocaloid:

I am not sure how well I can explain a) but I will try ;_; since all the points are given rigid transformations, the relative y- and x- distances between corresponding points of line s and r are the same, making the slopes equal something like that ;;

hardlyhuman:

Okay, let me try this.

Vocaloid:

for b) and c) you will just apply the slope formula (y2-y1)/(x2-x1) except using the variables from the points on lines r and s instead of numbers

hardlyhuman:

Okay, is there any way you could walk me through b and c

Vocaloid:

|dw:1515807756342:dw|

Vocaloid:

so (y2-y1)/(x2-x1) = (k-n)/(j-m) = your answer for c) it's just the same thing, except with the two points on line s

hardlyhuman:

Okay, thank you.

Vocaloid:

for d) since line q is a translation 3 units down of line s, just take the coordinates of point P' (make sure to use P' not P) and subtract 3 from the y-coordinate

hardlyhuman:

okay and for c) is it (y2-y1)/(x2-x1) = (k+e-n+e)/(j-j) ?

Vocaloid:

should be j - m not j - j

Vocaloid:

also be careful with your parentheses, the numerator needs to be (k+e) - (n+e)

hardlyhuman:

okay and with d) im kinda confused

Vocaloid:

so the coordinates of P' are (m,n+e) a shift of 3 units down means the y-coordinates are 3 less than they were before so just (m, n + e - 3) = your ans

hardlyhuman:

Okay, thank you. I'll send the last one

hardlyhuman:

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Vocaloid:

this website has a diagram that explains the construction better than I can ;_; https://www.mathsisfun.com/geometry/construct-linebisect.html

Vocaloid:

|dw:1515808608985:dw|

Vocaloid:

|dw:1515808613982:dw|

Vocaloid:

|dw:1515808634226:dw|

Vocaloid:

|dw:1515808682984:dw|

Vocaloid:

(keep the same width)

hardlyhuman:

So basically make it into a triangle?

Vocaloid:

|dw:1515808703887:dw|

Vocaloid:

then just connect the two places where the arcs intersect

Vocaloid:

|dw:1515808732767:dw|

Vocaloid:

I hope that makes sense, my brain is starting to get fried ;;

hardlyhuman:

yeah, is that the finished line or is there more i need to do to it

Vocaloid:

that's it

hardlyhuman:

kk last one

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Vocaloid:

first find the slope of y = -3x - 8

Vocaloid:

brb

hardlyhuman:

kk

hardlyhuman:

-3 is the slope

Vocaloid:

good, and it wants the perpendicular line, so we take the negative reciprocal to get (1/3) as the new slope then we re-write y = (1/3)x + b then we plug in (-3,1) as our new x and y values and solve for b

hardlyhuman:

awesome, thank you

hardlyhuman:

b=31x+y ?

Vocaloid:

so x = -3 and y = 1 y = (1/3)x + b 1 = (1/3)(-3) + b b = ?

hardlyhuman:

2?

Vocaloid:

yup so your final answer is y = (1/3)x + 2

hardlyhuman:

awesome, thank you so much for all your help. hope you have a good day/night :)

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