You can try them out and see which one works.
a: f(2) = f(1) +6 = 5+6 = 11 . . . . . . not this one
b: f(1) = f(2) -6 = -1-6 = -7 . . . . . . not this one (5 ≠ -7)
c: f(2) = f(1) - 6 = 5 - 6 = -1 . . . . . this gives the right f(2)
d: f(2 = -6(f(1) = -6(5) = -30 . . . . not this one
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The appropriate choice is ...
... f(n +1) = f(n) - 6
— — — — —
You can also recognize that the next term is 6 less than the current one, so f(n+1) = f(n) - 6, which corresponds to the 3rd selection.
Answer:
B) x = 7, y = 4
Step-by-step explanation:
you can find the value of 'y' first by connecting the bases with another altitute measuring 4 units
you now have an isosceles triangle where each leg is 4 which makes the hypotenuse equal to 4
to find 'x', it is the sum of 3 and 4
Answer:
i)W = 2500 / T
ii) W = 500 Tons
iii) grad W(10°) = - 25î
iv) The formulation is not practical
Step-by-step explanation:
i) Write an equation describing the use of coal
As use of coal is inversely proportional to the average monthly temperature
if W is use of coal in tons/per month then
W(t) = k / T where k is a constant of proportionality and T is the average temperature in degrees. We have to determine k from given conditions
k = ?? we know that when T = 25° W = 100 tons the by subtitution
W = k/T 100 = k /25 k = 2500 Tons*degree
Then final equation is:
W = 2500 / T
ii) Find the amount of coal when T = 5 degrees
W = 2500 / 5
W = 500 Tons
iii)
The inverse proportionality implies that W will decrease as T increase.
The vector gradient of W function is:
grad W = DW(t)/dt î
grad W = - 2500/T² î
Wich agrees with the fact that W is decreasing.
And when T = 100°
grad W(10°) = - 2500/ 100 î ⇒ grad W(10°) = - 25î
iv) When T = 0 The quantity of coal tends to infinite and the previous formulation is not practical
I think the answer is this
2x-4x+1=0
Combine like terms
-2x+1=0
Subtract 1 both sides
-2x=-1
Divide by -2 both sides
x=-1/2
