Answer:
The range is y ≥ -1
Step-by-step explanation:
∵ f(x) = (x - 4)(x - 2)
∴ f(x) = x² - 2x - 4x + 8
∴ f(x) = x² - 6x + 8 ⇒ quadratic function (ax² + bx + c) represented by
parabola graphically
∵ a = 1 , b = -6 , c = 8
∴ x-coordinate of its vertex = -b/2a = -(-6)/2×1 = 3
∴ f(3) = (3)² - 6(3) + 8 = -1
∵ a is positive ⇒ the curve has minimum point and it's open upward
∴ the minimum point is (3 , -1)
∴ The range is y ≥ -1 ⇒ because the minimum value is -1
note that gradient =
at x = a
calculate
for each pair of functions and compare gradient
(a)
= 2x and
= - 1
at x = 4 : gradient = 8 and - 1 : 8 > - 1
(b)
= 2x + 3 and
= - 2
at x = 2 : gradient = 7 and - 2 and 7 > - 2
(c)
= 4x + 13 and
= 2
at x = - 7 : gradient = - 15 and 2 and 2 > - 15
(d)
= 6x - 5 and
= 2x - 2
at x = - 1 : gradient = - 11 and - 4 and - 4 > - 11
(e)
y = √x = 
= 1/(2√x) and
= 2
at x = 9 : gradient =
and 2 and 2 > 
Answer:
1 solution
Step-by-step explanation:
-2(3h - 1) =5(2h - 3)
-6h + 2 = 10h - 15
-16h = -17
h = 17/16
Answer: 1 solution
Use Law of Cooling:

T0 = initial temperature, TA = ambient or final temperature
First solve for k using given info, T(3) = 42

Substituting k back into cooling equation gives:

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:

Solve for x:

Sub back into original cooling equation, x = T(t)

Solve for t:

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM