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1 answer:
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Neato
assuming you mean
remember
in form
a is leading coficient
(h,k) is vertex
k will be the minimum or max value of the function
if a is positive, then the parabola opens up and k is minimum
if a is negative, then the parabola opens down and k is max
given
a is negative
the parabola opens down and k is the max value
it is -3
the max value is -3, which occors at x=2
g(2)=-3 is max
assuming you mean
remember
in form
a is leading coficient
(h,k) is vertex
k will be the minimum or max value of the function
if a is positive, then the parabola opens up and k is minimum
if a is negative, then the parabola opens down and k is max
given
a is negative
the parabola opens down and k is the max value
it is -3
the max value is -3, which occors at x=2
g(2)=-3 is max
The sum of the lengths of three sides of the rectangle, gives the length
of the fencing, while one third of the rectangle area is for the pavement.
Responses:
Project A: The formula for the length of the fencing is, L = 4·x - 1
Project B: Length of the fancy wall = 2·x + 1
<h3>Which method can be used to find the length and area of paving from the given equations?</h3>
Project A: Let QP and SR represent the longest sides of the rectangle, we have;
PQ = SR = 2·x + 1
Given parameters are;
Length of the rectangular plot = 2·x + 1
Width of the rectangular plot = x - 1
Vertices of the rectangular plot are; QPSR
Project A: Let QP and SR represent the longest sides of the rectangle, we have;
PQ = SR = 2·x + 1
Which gives;
SP = QR = x - 1
The length of the fencing, L = SP + PQ + QR = x - 1 + 2·x + 1 + x - 1 = 4·x - 1
- The formula for the length of the fencing is, L =<u> 4·x - 1</u>
Project B: The front side is SR
Therefore;
- Length of the fancy wall = <u>2·x + 1</u>
Project C:
Area, A = Length × Width
Area of the plot, A = (2·x + 1) × (x - 1) = 2·x² - x - 1
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