Answer:
8.
10x-1=14x-37[corresponding side of congruent triangle are equal]
37-1=14x-10x
36=4x
x=36/4
x=9
9.
6x+13=16x-7[corresponding angle of congruent triangle are equal]
13+7=16x-6x
10x=20
x=20/10
x=2
10.
8x-27=4x+33
8x-4x=33+27
4x=60
x=60/4
x=15
PQ=8×15-27=93
11.3x+5=12x-13
13+5=12x-3x
9x=18
x=18/9=2
now
<BDC=12×2-13=11
12.
5x-8=9x-28
28-8=9x-5x
4x=20
x=20/4
x=5
now
<YWX=5x-8+9x-28=5×5-8+9×5-28=34
13..5x-4=2x+5
5x-2x=5+4
3x=9
x=9/3=3
now
ML=5×3-4=11
ML=11
Answer:
Step-by-step explanation:
Given the following vectors a = (-3,4) and b = (9, -1)
|a| and |b| are the modulus of a and b respectively.
|a| = √(-3)²+4²
|a| = √9+16
|a| = √25
|a| = 5
Similarly;
|b| = √(9)²+1²
|b| = √81+1
|b| = √82
We are to find the following;
a) a + b
a+b = (-3,4) + (9, -1)
a+b = (-3+9, 4+(-1))
a+b = (6, 4-1)
a+b = (6,3)
b) 8a + 9b
8a + 9b = 8(-3,4) + 9(9, -1)
8a + 9b = (-24,32) + (81, -9)
8a + 9b = (-24+81, 32+(-9))
8a + 9b = (57, 32-9)
8a + 9b = (57, 23)
c) |a| = √(-3)²+4²
|a| = √9+16
|a| = √25
|a| = 5
d) |a − b|
To get |a − b|, we need to get a-b first
Solve for a -b
a-b = (-3,4) - (9, -1)
a-b = (-3-9, 4-(-1))
a-b = (-12, 4+1)
a-b = (-12,5)
Find modulus of a-b i.e |a − b|,
|a − b| = √(-12)²+5²
|a − b| = √144+25
|a − b| =√169
|a − b| = 13
So you can do this multiple ways, I'll do this the way that I think makes sense the l most easily.
Cos (0) = 1
Cos (pi/2)=0
Cos (pi) =-1
Cos (3pi/2)=0
Cos (2pi)=1
Now if you multiply the inside by 4, the graph oscillates more violently (goes up and down more in a shorter period).
But you can always reduce it.
Cos (0)= 1
Cos (4pi/2) = cos (2pi)=1
Cos (4pi) =Cos (2pi) =1 (Any multiple of 2pi ==1)
etc...
the pattern is that every half pi increase is now a full period as apposed to just a quarter of one. That's in theory.
Now that you know that, the identities of Cosine are another beast, but mathematically.
You have.
Cos (2×2t) = Cos^2 (2t)-Sin^2 (2t)
Sin^2 (t)=-Cos^2 (t)+1..... (all A^2+B^2=C^2)
Cos (2×2t) = Cos^2 (t)-(-Cos^2 (t)+1)
Cos (2×2t)= 2Cos^2 (2t) - 1
2Cos^2 (2t) -1= 2 (Cos^2(t)-Sin^2(t))^2 -1
(same thing as above but done twice because it's cos ^2 now)
convert sin^2
2Cos^2 (2t)-1 =2 (Cos^2 (t)+Cos^2 (t)-1)^2 -1
2 (2Cos^2(t)-1)^2 -1
2 (2Cos^2 (t)-1)(2Cos^2 (t)-1)-1
2 (4Cos^4 (t) - 2 (2Cos^2 (t))+1)-1
Distribute
8Cos^4 (t) -8Cos^2 (t) +1
Cos (4t) =8Cos^4-8Cos^2 (t)+-1
Answer:
Second graph
Step-by-step explanation:
See attached image.
