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 >
If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.
Answer:
D. The limiting value for the height of the plant is 6.88 cm.
Step-by-step explanation:
Answer:
The slope of the line is .
Step-by-step explanation:
We are given two coordinate points:
We are asked to find the slope of the line.
We can use the rise-over-run formula to solve for the slope of the line.
However, we firstly need to name our coordinate points.
In math, we can label our coordinates using the following label system:
Therefore, we can also label our coordinates as such:
Now, we can supply these values into the formula and solve for our slope, or a better known variable, <em>m</em>.
Therefore, our slope is .