The domain is all real numbers
The Range is y > 0
AS x increases by 1 the value of y is 1/7 of previous value
So
y=ax^2+bx+c
(x,y)
sub the points and solve
(4.28,6.48)
6.48=a(4.28)^2+b(4.28)+c
(12.61,15.04)
15.04=a(12.61)^2+b(12.61)+c
well, for 3 variables, we need equations and therefor 3 points
maybe we are supposed to assume it starts at (0,0)
so then
0=a(0)^2+b(0)+c
0=c
so then
6.48=a(4.28)^2+b(4.28)
15.04=a(12.61)^2+b(12.61)
solve for a by subsitution
first equation, minut a(4.28)^2 from both sides
6.48-a(4.28)^2=b(4.28)
divide both sides by 4.28
(6.48/4.28)-4.28a=b
sub that for b in other equation
15.04=a(12.61)^2+b(12.61)
15.04=a(12.61)^2+((6.48/4.28)-4.28a)(12.61)
expand
15.04 =a(12.61)^2+(81.7128/4.28)-53.9708a
minus (81.7128/4.28) both sides
15.04-(81.7128/4.28)=a(12.61)^2-53.9708a
15.04-(81.7128/4.28)=a((12.61)^2-53.9708)
(15.04-(81.7128/4.28))/(((12.61)^2-53.9708))=a
that's the exact value of a
to find b, subsitute to get
(6.48/4.28)-4.28((15.04-(81.7128/4.28))/(((12.61)^2-53.9708)))=b
if we aprox
a≈-0.038573167896199
b≈1.6791118501845
so then the equation is
y=-0.038573167896199x²+1.6791118501845x
The given figure shows a vertical hyperbola with its centre at origin, and as we observe the figure, we can conclude that :
Length of transverse axis is :
length of conjugate axis is :
Equation of hyperbola ~
plug in the values ~
Answer:
The length of AA' = √29 = 5.39
Step-by-step explanation:
* Lets revise how to find the length of a line joining between
any two points in the coordinates system
- If point A is (x1 , y1) and point B is (x2 , y2)
- The length of AB segment √[(x2 - x1)² + (y2 - y1)²]
* Lets use this rule to solve the problem
∵ Point A is (0 , 0)
∵ Point A' = (5 , 2)
∵ (x2 - x1)² = (5 - 0)² = 5² = 25
∵ (y2 - y1)² = (2 - 0)² = 2² = 4
∴ The length of AA' = √(25 + 4) = √29 = 5.39
Answer: 30
Step-by-step explanation:
