Firstly, It says that the two triangles are similar .
The scale factor is 2 as 12 / 6 = 2
25 + 85 = 110
180 - 110 = 70.
70 degrees
Δ = 1 + 4*20 = 81
x' = (-1 + 9)/2 = 4
x'' = (-1 - 9)/2 = -5
4² + 4 - 20 = 0
(-5)² - 5 - 20 = 0
Ok !
8/9 = 0,8
0,8/4 = 0,2
so each piece was 0,2 yard long
Answer:
a) Var[z] = 1600
D[z] = 40
b) Var[z] = 2304
D[z] = 48
c) Var[z] = 80
D[z] = 8.94
d) Var[z] = 80
D[z] = 8.94
e) Var[z] = 320
D[z] = 17.88
Step-by-step explanation:
In general
V([x+y] = V[x] + V[y] +2Cov[xy]
how in this problem Cov[XY] = 0, then
V[x+y] = V[x] + V[y]
Also we must use this properti of the variance
V[ax+b] =
V[x]
and remember that
standard desviation = ![\sqrt{Var[x]}](https://tex.z-dn.net/?f=%5Csqrt%7BVar%5Bx%5D%7D)
a) z = 35-10x
Var[z] =
Var[x] = 100*16 = 1600
D[z] =
= 40
b) z = 12x -5
Var[z] =
Var[x] = 144*16 = 2304
D[z] =
= 48
c) z = x + y
Var[z] = Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80
D[z] =
= 8.94
d) z = x - y
Var[z] = Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80
D[z] =
= 8.94
e) z = -2x + 2y
Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320
D[z] =
= 17.88
Answer:
Part 1) The domain of the quadratic function is the interval (-∞,∞)
Part 2) The range is the interval (-∞,1]
Step-by-step explanation:
we have

This is a quadratic equation (vertical parabola) open downward (the leading coefficient is negative)
step 1
Find the domain
The domain of a function is the set of all possible values of x
The domain of the quadratic function is the interval
(-∞,∞)
All real numbers
step 2
Find the range
The range of a function is the complete set of all possible resulting values of y, after we have substituted the domain.
we have a vertical parabola open downward
The vertex is a maximum
Let
(h,k) the vertex of the parabola
so
The range is the interval
(-∞,k]
Find the vertex

Factor -1 the leading coefficient

Complete the square


Rewrite as perfect squares

The vertex is the point (7,1)
therefore
The range is the interval
(-∞,1]