Answer: this is a continuation starting in number 4 (use the same steps as example 3)
Step-by-step explanation:
-12
45
-12
14
32
-2
5
Answer:
Step-by-step explanation:
Question says that it uses 120 feet of fencing material to enclose three sides of the play area. This means there are 3 sides. Putting this into equation, we have something like this.
120 = L + 2W
Where
LW = area.
Again, in order to maximize the area with the given fencing, from the equation written above, then Width, w must be = 30 feet and length, l must be = 60
On substituting, we have
A = LW = (120 - 2W) W
From the first equation, making L the subject of the formula, we have this
L = 120 - 2W, which then we substituted above.
On simplification, we have
L = 120W -2W²
Differentiating, we have
A' = 120 - 4W = 0
Remember that W = 30
So therefore, L = 120 - 2(30) = 60 feet
This gives you three simultaneous equations:
6 = a + c
7 = 4a + c
1 = c
<u>c = 1
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If c =1,
6 = a + 1
<u>a = 5
</u><u /><u />
This doesn't work in the second equation, so the quadratic that goes through these points is not in the form y = ax^2 + bx + c
Was there supposed to be a b in the equation?
Answer:
w = 0 , -3 , 3
Step-by-step explanation:
w³ - 9w = 0
w(w² - 9) = 0
w(w² - 3²) = 0
w(w+3)(w-3) = 0 {a² - b² = (a + b)(a - b) }
w = 0 ; w + 3 = 0 ; w - 3 = 0
w = -3 ; w = 3
w = 0 , -3 , 3
Slope for (x₁, y₁) and (x₂, y₂): (6, 2) and (7, 4)
Slope: (y₂ -y₁) / (x₂ -x₁)
Slope: ( 4 - 2) / (7 - 6) = 2/1
<span>Slope = 2/1 = 2
Option D</span>
