Length = w+3
P = 2L +2W
50 = 2(w+3) + 2w
50 = 2w +6 +2w
50 = 4w +6
50-6 =4w-6
44 = 4w
44/4 = 4w/4
w = 11
l = 11+3
l = 14
50 = 2L+2W
50 = 2(14) + 2(11)
50 = 28 +22
50=50
Answer:
Step-by-step explanation:
8 2
16 4
24 6
32 8
40 10
???
Answer:
[1, 1]
Step-by-step explanation:
Translation → [-1, 3] moves down to [-1, 1]
Now, a <em>90°-clockwise rotation</em> is the exact same as a <em>270°-counterclockwise rotation</em>, and according to the <em>270°-counterclockwise rotation</em> [<em>90°-clockwise rotation</em>] <em>rule</em>, you take the y-coordinate, bring it over to your new x-coordinate, and take the OPPOSITE of the x-coordinate and set it as your new y-coordinate:
<u>Extended Rotation Rules</u>
- 270°-clockwise rotation [90°-counterclockwise rotation] >> (<em>x, y</em>) → (<em>-y, x</em>)
- 270°-counterclockwise rotation [90°-clockwise rotation] >> (<em>x, y</em>) → (<em>y, -x</em>)
- 180°-rotation >> (<em>x, y</em>) → (<em>-x, -y</em>)
Then, you perform your rotation:
270°-counterclockwise rotation [90°-clockwise rotation] → [-1, 1] moves to [1, 1]
I am joyous to assist you anytime.
Answer:
Explanation:
Given:
The equation describing the forest wood biomass per hectare as a function of plantation age t is:
y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4
The equation that describes the annual growth in wood biomass is:
y ′ (t) = 0.01t + 0.072t^2 - 0.018t^3
To find:
a) The year the annual growth achieved its highest possible value
b) when does y ′ (t) achieve its highest value?
a)
To determine the year the highest possible value was achieved, we will set the derivative y'(t) to zero. The values of t will be substituted into the second derivative to get the highest value
SInce t = 4.13, gives y ′' (t) = -0.316 (< 0). This makes it the maximum value of t
The year the annual growth achieved its highest possible value to the nearest whole number will be
year 4
b) y ′ (t) will achieve its highest value, when we substitute the value of t that gives into the initial function.
Initial function: y(t) = 5 + 0.005t^2 + 0.024t^3 − 0.0045t^4