Answer:
Step-by-step explanation:
<u>The transformations include:</u>
This is a dilation by a scale factor of 3 and then translation 2 units left and 2 units up.
<u>The transformation applied to the point U:</u>
- U(-4,2) → U'(-4*3 - 2, 2*3 + 2) = U'(-14, 8)
Answer:
a carrot? yea a carrot. :D
Answer:
First, let's make an expression!
If you meant a different equation..sorry
![\frac{8+6}{7-3} +\frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B8%2B6%7D%7B7-3%7D%20%2B%5Cfrac%7B1%7D%7B2%7D)
Hope this is what you meant!
![\frac{14}{4} +\frac{1}{2}\\\frac{7}{2} +\frac{1}{2} \\\frac{8}{2} =4](https://tex.z-dn.net/?f=%5Cfrac%7B14%7D%7B4%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%5C%5C%5Cfrac%7B7%7D%7B2%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5C%5C%5Cfrac%7B8%7D%7B2%7D%20%3D4)
Hope this helps!
P.S. Stay Safe!
Note that
![\displaystyle{ \sin30^{\circ}= \frac{1}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20%5Csin30%5E%7B%5Ccirc%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20)
and
![\displaystyle{ \cos30^{\circ}= \frac{ \sqrt{3}}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20%5Ccos30%5E%7B%5Ccirc%7D%3D%20%5Cfrac%7B%20%5Csqrt%7B3%7D%7D%7B2%7D%20)
.
From the unit circle, we can check that
![\sin30^{\circ}](https://tex.z-dn.net/?f=%5Csin30%5E%7B%5Ccirc%7D)
and
![\sin(-30)^{\circ}](https://tex.z-dn.net/?f=%5Csin%28-30%29%5E%7B%5Ccirc%7D)
have the same value, but opposite signs, that is:
![\sin(-30)^{\circ}=-\sin(30)^{\circ}=-\frac{1}{2}](https://tex.z-dn.net/?f=%5Csin%28-30%29%5E%7B%5Ccirc%7D%3D-%5Csin%2830%29%5E%7B%5Ccirc%7D%3D-%5Cfrac%7B1%7D%7B2%7D)
.
Thus,
![\displaystyle{ \cos(x-y)= \cos(-30^{\circ}-60^{\circ})=cos(-90^{\circ})](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%20%5Ccos%28x-y%29%3D%20%5Ccos%28-30%5E%7B%5Ccirc%7D-60%5E%7B%5Ccirc%7D%29%3Dcos%28-90%5E%7B%5Ccirc%7D%29)
.
Note that
![cos(-90^{\circ})](https://tex.z-dn.net/?f=cos%28-90%5E%7B%5Ccirc%7D%29)
is the x-coordinate of the lowest point on the unit circle, that is (0, -1). Thus
![cos(-90^{\circ})=0](https://tex.z-dn.net/?f=cos%28-90%5E%7B%5Ccirc%7D%29%3D0)
Answer: 0
1) Linear model
R(t) = y = at + b
Where t is the year - 2000 (year since 2000)
For year 2006, t = 6; for year 2010, t = 10
Then:
1) 10.7 = a*6 + b
2) 34.2 = a*10 + b
Subtract (1) from (2)
34.2 - 10.7 = 10a -6a
23.5 = 4a
a = 23.5/4 = 5.875
Now from (1) 10.7 = 6a + b => b = 10.7 - 6a = 10.7 - 6*5.875 = - 24.55
Then the resulting model is R(t) = 5.875a - 24.55
2) Exponential model
R(t) = A[B]^t
(1) 10.7 = A[B]^6
(2) 34.2 = A[B]^10
Divide (2) by (1)
[34.2/10.7] = [B]^10 / [B]^6
3.1963 = [B]^4 => B = 1.3371
Now, from (1) 10.7 = A [1.3371]^6 => A = 10.7 / [1.337]^6 = 1.8725
Then, the model is R(t) = 1.8725{1.3371]^t