Answer:
Generally it's one that is just out of the way, the extreme outcast.
Answer:
The answer is 20
Step-by-step explanation:
From the graph it shows that J is located at point 15, and Q is located at point 35
If we move 20 units from point J to point Q, we land at point Q, making it the distance they are from one another.
A
given the 2 equations
2x + y = - 2 → (1)
x + y = 5 → (2)
subtract (2) from (1) term by term to eliminate y
(2x - x ) + (y - y ) = (- 2 - 5 )
x + 0 = - 7 ⇒ x = - 7
substitute x = - 7 in either of the 2 equations and solve for y
(2) : - 7 + y = 5 ( add 7 to both sides )
y = 5 + 7 = 12 → A
Answer:
0.054
Step-by-step explanation:
Using the compound formula A=P(1+r/n)^nt we can plug in the numbers we know. A being the total, P the initial amount, n is the number of compounds in a year, and t is the number of years passing. So 10300= 4900(1+ r/2)^2(14). We have to isolate r in order to find the rate. We can divide both sides by 4900, giving us 103/49= (1+ r/2)^28. Then we take the 28 square root from both sides. 1.02689= 1 + r/2. Subtract 1 from both sides. 0.02689= r/2. Then multiply both sides by two. r= 0.054.
Answer: f(x) = -0.016x² + 1.6x
Step-by-step explanation:
If the lenght of the road over the arch is 100m, we can consider a coordinate plane and say that the road starts at point (0,0) and finishes at (100,0). The vertice of the parabola is at point (50,40), because the maximum height is 40 and it is always in the middle point of the roots.
So, we have
(0,0) (50,40) (100,0)
A quadratic function is always on the form: f(x) = ax² + bx + c
0 = a0² + b0 + c
40 = a50² + b50 + c
0 = a100² + b100 + c
0 = a0² + b0 + c → c = 0 ∴
40 = a50² + b50
0 = a100² + b100
_________________________
2500a + 50b = 40 (*2)
10000a + 100b = 0
_________________________
5000a + 100b = 80
10000a + 100b = 0 (-)
__________________________
-5000a = 80
-a=80/5000
a=-0.016
∴
2500a + 50b = 40
2500.(-0.016) + 50b = 40
-40 + 50b = 40
50b = 80
b = 80/50
b = 1.6
This way f(x) = -0.016x² + 1.6x
