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sammy [17]
3 years ago
8

HELP PLEASE NEED BEFORE TOMORROW MORNING

Mathematics
1 answer:
9966 [12]3 years ago
5 0
Okay YAS, I loove linear equations... moving on. 

OKAY, so I'm going to start with Q2 (not 100% sure how to work out 1, but lets use process of elimination). Firstly, the rule is y=mx+c So, start with the first line y=2x-3, from this the first thing you need to do is find the y-intercept (m) and the gradient (c). So, this is very simple 2 is m and -3 is c (if you look at the rule it matches up). 
m=2/1(this is rise/run, rise is how much you go up by, and run is how much you go left or right by (left if positive, right if negative))
c=-3
Now, go to your graph, and look on the y line and go to -3. Then go left, 1, and up 2. IS THERE A LINE CROSSING THROUGH THIS POINT? Yes? Good. 
Now for y=-1/2+1. Looking at the rule again (y=mx+c), 
m=-1/2 (rise/run)
c=1
NOW, once again go to your graph, and look on the y line and go to 1 (in this case the line actually passes through this point), now go right 2 places and up 1. IS THERE A LINE? Yes? Good!

NOOOW, for Q3, use the same steps and you will have the answer. 
y=1/4x-2
m=1/4 (rise/run)
c=-2
Look at graph C, start at -2 on the y line, go left 4, and up 1? Yes? Line crossing through this point? good!!
Now for the 2nd one, 
y= -2/5x+4
m=-2/3
c=4
Look at the same graph start at 4 on the y line, go right 5 and up 2! Got it? There's a line crossing through this point, yes? YAY

Now I have no idea how to do Q1, but by process of elimination it's pretty obvious. I hope this is right! Okay, bye!! :D
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Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has
Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving 

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

<span>AB = 6 units   BC = 11 units AC = 12.53 units
</span><span>∆XYZ  X(-10, 5), Y(-12, -2), and Z(-4, 15)
</span><span>XY = 7.14 units   YZ = 18.79 units XZ = 11.66 units</span>

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>MN = 6 units   NO = 11 units MO = 12.53 units
</span><span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).
</span><span>JK = 5 units   KL = 9 units JL = 10.30 units
</span><span>∆PQR  P(12, 3), Q(12, -2), and R(3, -2)
</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span> 
Therefore
<span>we have the <span>∆ABC   and the </span><span>∆MNO  </span><span> 
with all three sides equal</span> ---------> are congruent  
</span><span>we have the <span>∆JKL  </span>and the <span>∆PQR 
</span>with all three sides equal ---------> are congruent  </span>

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

 1)     If ∆MNO   ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC 

<span> then </span>∆MNO<span> ≅</span> <span>∆ABC  </span> 

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)----------------->  M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)----------------->  M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)----------------->  M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

<span>Therefore ∆MNO ≅ <span>∆ABC   - > </span>check list</span>

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

<span> then </span>∆JKL ≅ <span>∆PQR  </span> 

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)----------------->  J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

<span>∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)----------------->  J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

<span>Therefore ∆JKL ≅ <span>∆PQR   - > </span><span>check list</span></span>
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