Answer:
see explanation
Step-by-step explanation:
(1)
Given
g(r) = (r + 14)² - 49
To obtain the zeros, let g(r) = 0 , that is
(r + 14)² - 49 = 0 ( add 49 to both sides )
(r + 14)² = 49 ( take the square root of both sides )
r + 14 = ± 
= ± 7 ( subtract 14 from both sides )
r = - 14 ± 7, then
r = - 14 - 7 = - 21 ← smaller r
r = - 14 + 7 = - 7 ← larger r
(2)
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
g(r) = (r + 14)² - 49 ← is in vertex form
with vertex = (- 14, - 49 )
Answer:
E
Step-by-step explanation:
If you add 22 onto the sum of all the points and divide by the amount of games (6), you get 12.
Answer:
- see below for a drawing
- the area of one of the trapezoids is 20 units²
Step-by-step explanation:
No direction or other information about the desired parallelogram is given here, so we drew one arbitrarily. Likewise for the segment cutting it in half. It is convenient to have the bases of the trapezoids be the sides of the parallelogram that are 5 units apart.
The area of one trapezoid is ...
A = (1/2)(b1 +b2)h = (1/2)(3+5)·5 = 20 . . . . square units
The sum of the trapezoid base lengths is necessarily the length of the base of the parallelogram, so the area of the trapezoid is necessarily 1/2 the area of the parallelogram. (The area is necessarily half the area of the parallelogram also because the problem has us divide the parallelogram into two identical parts.)
Part A:
y=mx+b
621=83(7)+b
Part B:
621=83(7)+b
621=581+b
621-581=b
b=40
Insurance cost is 40$
Simple :)
Area of shaded part = are of 1/4 circle - area of both triangles.
Are of circle = pie r^2 so 100x3.14 = 314 cm2.
Area of triangle AOB= area of triangleDOE = bh/2= 5x10/2= 25 each
However, the traingles share a common area which is quad DOB(I)
Lets take traingle AOE, whose are is bh/2=10x10/2=50cm2.
50-area of triangle A(I)E= 50-(
