tw101
  • tw101
Please help! I have an assignment that I have been trying to do but i'm a bit stuck. I have done the first questions but need help on the last ones. Help would be much appreciated. (i will put pictures below)
Mathematics
chestercat
  • chestercat
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tw101
  • tw101
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tw101
  • tw101
This is what I have. And I have already done questions 1,2,and 3, but only got halfway on 6&7. I drew the triangle and rectangle but I'm not sure how to get the length of the sides.Im also not sure how to do 4&5. I know that for parallel the slopes are the same and for perpendicular they're opposite reciprocals, but do I have to draw two new lines to accommodate the given points?
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phi
  • phi
to do Q4, we need to know the equation of the line for Q2 (for example) a parallel line will have the same slope I don't know what slope you got for Q2, but pretend it was 2. then the equation for Q4 would be y = 2x + b (same slope, but we need to find b) they say this line goes through the point (1,-3) (i.e. when x is 1, y is -3) so use that info to get -3 = 2*1 + b or -3 = 2+ b to find b, add -2 to both sides and simplify but remember we need the slope for Q2 (which is probably not 2)

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phi
  • phi
for Q5 a perpendicular slope to slope "m" is -1/m for example if m is 2/3 then the perpendicular slope is -3/2 (flip and negate) or: -1/2 and 2 or 2 and -1/2 or 3 and -1/3 say you choose the line from Q2. get its slope and "flip it" then multiply by -1 after you get the slope, it is the same problem as Q4... you have to find b
phi
  • phi
for Q6, I would pick a "simple triangle" (a right triangle with legs that go straight across and straight up/down) we don't have to, but it is easier. the length of the straight across line is easy: count the number of squares (or subtract the x values) for a slanted line you have to use the distance formula do you have that in your notes?
phi
  • phi
I don't know if you have to plot the lines for Q4 or Q5 but it can't hurt. once you have the equation, find two points on the new line and plot them... then connect the dots
tw101
  • tw101
THank you so much for helping me! sorry i didnt respond right away! I didnt realize that someone was actually looking at my question. I had given up haha.
phi
  • phi
try to make progress if you post your work someone (me? if I'm here) will check it.
tw101
  • tw101
okay. for 4 and 5 what do you mean by plot lines?
phi
  • phi
you wrote *** but do I have to draw two new lines to accommodate the given points?*** which I assumed meant: do I have to plot the new lines which will be parallel or perpendicular to the original line? and I thought it would not hurt.
tw101
  • tw101
oh okay, I see.
tw101
  • tw101
I was going to do the line I drew over the left side of the triangle since its goes through the points it gave me. would that work?
phi
  • phi
what two points does it go through. I'm not sure which line you mean
tw101
  • tw101
one second. Ill show the updated picture
tw101
  • tw101
I cut the triangle in half to make it how you said it would be easier.
tw101
  • tw101
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tw101
  • tw101
the line that runs alongside the left side of the triangle
phi
  • phi
ok, but first we have to fix that line. it goes through (2,5) and (4,8) its slope is change in y divided by change in x (8-5)/(4-2) or 3/2 we can say y = (3/2)x + b now use (2,5): 5 = (3/2)*2 + b 5= 3+b b= 2 and the equation is y= (3/2)x + 2
tw101
  • tw101
okay, so where would I go from there
phi
  • phi
notice, when y= 19 (top of the outer box) 19= (3/2)x + 2 add -2 to both sides 17 = 3/2 x multiply both sides by 2/3 34/3 = x or x = 11.333 (11 and 1/3) not x=12 as you have. in other words your line on the graph is a little off
tw101
  • tw101
okay, so would i move it up just a little?
phi
  • phi
not up. I would move the point at (12,19) to (11 1/3 , 19) you will have to estimate 1/3 of box over from 11 then redraw the line to go through the 3 points (2,5), (4,8) and (11 1/3, 19)
tw101
  • tw101
Okay. The (4,8) is the lower left part of the rectangle..
phi
  • phi
ok. after you fix the line we can do Q4
tw101
  • tw101
Okay:)
phi
  • phi
btw, y= (3/2) x + 2 will be an answer for Q2 . a line with positive slope and you have the work up above on how we found the equation
phi
  • phi
Q4 ... the question is a bit muddled but I assume it is the new line that goes through (1,-3) and they want it parallel to y= (3/2) x + 2 parallel means the same slope. so write the new equation. can you do that ?
tw101
  • tw101
so would it just be y=(1/-3)x+2?
phi
  • phi
start with y = m x + b (the "skeleton" that we fill in) we want this line to be parallel to y= (3/2)x + 2 what slope should we use for m ? (parallel means *same* slope)
tw101
  • tw101
okay, so would the (3/2) be the same then?
phi
  • phi
Q1 is making the graph. you mean Q2 and Q3? for the "old line" we can use the line you used for Q2 what was it ?
tw101
  • tw101
yes, i already did 1,2, and 3
tw101
  • tw101
You have to choose 2 locations that make a line and are either positive or negative. I
phi
  • phi
what did you choose?
tw101
  • tw101
i chose (2,5) (12,19) for positive (21,5) (12,19) for negative.
tw101
  • tw101
so for Q4, would it still be y=(3/2) +b
tw101
  • tw101
but we still have to find b
phi
  • phi
if you change to (2,5), (4,8) we will get 3/2 for the slope (a much nicer number than if we use (12,19)
tw101
  • tw101
okay
phi
  • phi
and change (21,5) (12,19) for negative. to (21,5), (19, 8) (same reason..a nicer slope)
tw101
  • tw101
make both of them negative?
phi
  • phi
no. for Q2 use the points (2,5) and (4,8) (slope will be 3/2) for Q3 use the points (21,5), (19,8) (slope will be -3/2)
tw101
  • tw101
Ohhh okay haha
phi
  • phi
I hope that gets straightened out. but for Q4, we want a line parallel to the line from Q1 which has a slope of 3/2 we start with y= mx + b we replace m with 3/2 y= (3/2) x + b we want this new line to go through (1,-3) replace x and y with those numbers -3 = (3/2)*1 + b can you find b ?
tw101
  • tw101
would i solve for b
phi
  • phi
yes. can you do that?
tw101
  • tw101
I mean do i isolate it
tw101
  • tw101
yes
tw101
  • tw101
okay so i got (-3/1)-(3/2)
phi
  • phi
yes, but people would combine that either put -3 over the common denominator of 2 (i.e. write it as -6/2 ) or change 3/2 to 1.5 and figure out -3 - 1.5
tw101
  • tw101
okay so it would be -4.5?
phi
  • phi
yes, or -9/2 (same thing)
phi
  • phi
so the answer for Q4 is y = (3/2) x - 9/2
tw101
  • tw101
y=(3/2)*1+(-9/2)?
tw101
  • tw101
ok you already said that.
tw101
  • tw101
sorry haha
tw101
  • tw101
right, sorry
tw101
  • tw101
that was disgusting
tw101
  • tw101
wow. okay. im sorry you guys had to go through that.
tw101
  • tw101
i blocked them
phi
  • phi
ready for Q5? let's use the same line with slope (3/2) and we want a perpendicular line y= mx + b any idea what we use for m ?
tw101
  • tw101
(-3/2)?
tw101
  • tw101
since theyre reciprocals?
phi
  • phi
we want "flip and negate" you negated but did not flip
tw101
  • tw101
(-2/3)
phi
  • phi
yes so we have y = (-2/3) x + b and they want this to go through point (-2,7)
phi
  • phi
to find b, replace x and y with those numbers
tw101
  • tw101
okay so it would be 7=(-2/3)*-2+b
phi
  • phi
yes
tw101
  • tw101
then solve for b again right
phi
  • phi
yes
tw101
  • tw101
7/-2- (-2/3)=b
tw101
  • tw101
sorry i didnt put parenthesis on the first pair
phi
  • phi
that looks a bit peculiar. start with 7=(-2/3)*-2+b can you do -2 * -2/3 ?
tw101
  • tw101
(-4/6)?
tw101
  • tw101
(4,6)
tw101
  • tw101
sorry forgot a negative times a positive for a second. haha
phi
  • phi
it is \[ -2 \cdot \frac{-2}{3} = \frac{-2}{1} \cdot \frac{-2}{3}\] when you multiply fractions you multiply top * top and bottom * bottom
tw101
  • tw101
4/3?
phi
  • phi
yes. so you have 7 = 4/3 + b now add -4/3 to both sides
tw101
  • tw101
(7/-2)+(4/3)=b?
tw101
  • tw101
wait whoops
tw101
  • tw101
hold on, sorry
tw101
  • tw101
7+4/3? i was looking at my old equation
phi
  • phi
close. but added +4/3 we want -4/3 (because on the right side we want 4/3 - 4/3 which cancels)
tw101
  • tw101
im sorry, thank you for being so patient with me. math really isnt my stronest subject... at all.
tw101
  • tw101
okay, gotcha
tw101
  • tw101
so just 7+(-4/3)
phi
  • phi
b= 7 - 4/3 change 7/1 to 21/3 (multiply top and bottom by 3)
phi
  • phi
and yes you can write it 7 + (-4/3)
tw101
  • tw101
so 25/3?
tw101
  • tw101
21/3+4/3=25/3
phi
  • phi
if you did 21/3 + 4/3 we have 21/3 - 4/3
tw101
  • tw101
oh right ints negative
tw101
  • tw101
17/3
phi
  • phi
and the equation is ?
tw101
  • tw101
y=(-2/3)x+17/3?
phi
  • phi
yes
tw101
  • tw101
okay great
phi
  • phi
for Q6, did you change the top point from (12,19) to (11 1/3 , 19) ?
tw101
  • tw101
of the triangle?
phi
  • phi
yes (otherwise you will get marked off (if they notice!) because (12, 19) is not really on the line from (2,5) to (4,8)
tw101
  • tw101
yep
tw101
  • tw101
just draw a side straight down from that point now
phi
  • phi
ok, now figure out the slope of the straight up/down line can you do that ?
tw101
  • tw101
i think so
tw101
  • tw101
jsut a minute:)
tw101
  • tw101
okay so the two points are (11 1/3,19) and (11 1/3,4) so then i did the y1-y2/x1-x2 thing and got (0,15) not sure what to do now. would i substitute that into the equation? y=mx+b
phi
  • phi
they don't want the equation of the line. they want its slope change in y divided by change in x notice you will get (19-4)/ (11.333- 11.333) = 15/0 and divide by 0 is not allowed. we say the slope of a vertical line is undefined.
phi
  • phi
they also want the length of that side. it is 19-4 or 15
tw101
  • tw101
okay so slope:undefined length:15
phi
  • phi
yes. now do the bottom side
tw101
  • tw101
ok
phi
  • phi
though in your picture it looks like you only go down to y=5 (not y=4)
phi
  • phi
I would try to use the point (2,5) in this triangle because we know the slope of the slant side will be 3/2 (we did that work up above)
tw101
  • tw101
youre right. 14 then
tw101
  • tw101
sorry my internet is being really weird
phi
  • phi
yes. and change the number where you show your work.
tw101
  • tw101
okay so for the bottom side i got (11 1/3,5) (2,5) then when subtracted i got -1/7 1/3
phi
  • phi
ok, now you can do the bottom line (2,5) to (11 1/3, 5) change in y is (5-5) change in x is 11 1/3 - 2 = 9 1/3 and 0/ 9.333 is 0
tw101
  • tw101
ugh no that is not what i got
phi
  • phi
if you have a calculator, it will help
tw101
  • tw101
sorry my computer is freezing up
phi
  • phi
the length will be 9 1/3 (9 and 1/3) slope will be 0
phi
  • phi
as a check, you can count how long it is. but they want you to "show the work" so you should also show 11 1/3 - 2 = 9 1/3
tw101
  • tw101
would the slope be 0?
tw101
  • tw101
oh okay you said that haha sorry, again , slow computer
phi
  • phi
yes , slope is 0 (horizontal lines have slope 0)
phi
  • phi
now the slant side (2,5) to (11 1/3, 19) first find change in y and change in x
phi
  • phi
hint we did this already (see length of the vertical and horizontal sides)
tw101
  • tw101
yes, sorry i had to reset my connection
phi
  • phi
you should get 14 and 9 1/3
tw101
  • tw101
yes, i did. im so sorry idk what going on with my wifi
tw101
  • tw101
change of x 19-5 change of y 11 1/3-2
tw101
  • tw101
14 and 9 1/3 yes.
phi
  • phi
ok on your numbers, but x goes sideways... change in x is 9 1/3 the slope will be 14 / (9 1/3) to do that, write 9 1/3 as an improper fraction 28/3 when you divide by a fraction, you can "flip it" and multiply so 14 * 3/28 which is 3/2
tw101
  • tw101
ok
phi
  • phi
slope is 3/2 length we use the distance formula \[ \sqrt{ (y_2-y_1)^2 + (x_2-x_1)^2 } \] which is \[ \sqrt{ 14^2 + 9.3333^2} = \sqrt{196+87.11}= \sqrt{283.11}= 16.8\]
phi
  • phi
because the triangle has a right angle formed by the vertical and horizontal lines, it is a right triangle
tw101
  • tw101
oh okay i see.
phi
  • phi
for Q7 I would use the "mall" as the quadrilateral the slopes will be 0 for the top and bottom (horizontal lines) and undefined for the sides the lengths are easy (I hope... count how long each side is) because the shape has 4 90 degree angles and the opposite sides are equal it is a rectangle.
phi
  • phi
I have to go.
tw101
  • tw101
9 ad 15

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