Answer:
-1/4x + 3
Step-by-step explanation:
Answer:
Option B
Step-by-step explanation:
Taking the derivative of the residual function, then applying the normal equation:
[a; b] = (X'*X)^(X'*y)
with
X = [1 1; 2 1; 3 1; 4 1; 5 1]
X' is transpose of X
y = [11; 8; 4; 1; 0]
[a; b] = [-2.9 13.5]
Answer:
By AA
ΔWXY ~ΔWVZ
Step-by-step explanation:
Here WXY is an isosceles triangle with legs WX & WY
So WX = WY
Hence ∠X = ∠Y
So ∠2= ∠3.
Now by angle sum property
∠1 + ∠2+∠3 = 180°
∠1+∠2+∠2=180°
2∠2 = 180° - ∠1 .......(1)
In triangle WVZ
WV = WZ
So ∠V = ∠Z
∠4 = ∠5
Once again by angle sum property
∠1 + ∠4 + ∠5=180°
∠1 + ∠4 + ∠4 = 180°
2∠4 = 180° - ∠1 ...(2)
From (1) & (2)
2∠2 = 2∠4
∠2=∠4
Now ∠W is common to both triangles
Hence by AA
ΔWXY ~ΔWVZ
27)
Equation of the existing water pipe's line
a) slope, m = rise / run = Δy / Δx = 3/2
b) y-intercept, b = 3
equation: y = mx + b = (3/2)x + 3
Equation of the new water pipe's line
slope, m = - 1 / (slope of the perpendicular line) = - 1 / (3/2) = - 2/3
point (0,2)
=> y - 2 = (-2/3) (x - 0) => y = (-2/3)x + 2 <---- equation of the new pipe
28)
two parallel lines have the same slope =>
slope, m = rise / run = Δy / Δx = [4 - 0] / [11 - 8] = 4 / 3
point (4,5) => y - 5 = (4/3) (x - 4)
=> y = (4/3)x - 16/3 + 5 = (4/3)x - 1/3
y = (4/3)x - 1/3 <--- equation of the new bike path
