Which equations and/or functions represent the graphed line? Check all that apply.
2 answers:
The picture in the attached figure
step 1
find the equation of the line of the graph
let
A (-4,0) B(0,2)
slope m=(y2-y1)/(x2-x1)------> m=(2-0)/(0+4)----> m=1/2----> m=0.5
with m=0.5 and point B (0,2)
y-y1=m*(x-x1)-----> y-2=(0.5)*(x-0)-----> y=0.5x+2
step 2
find the solution
<span>
a) f(x) = 0.2x - 4----> the </span>function not represent the graphed line<span>
b) f(x) = 0.5x + 2----> </span> the function represent the graphed line<span>
c) f(x) = x + 2</span>--> the function not represent the graphed line<span>
d) y – 3 = (x – 2)---> y=x+1---> </span>the function not represent the graphed line<span>
e) y – 1 = 0.5(x + 2)</span>-----> y=0.5x+1+1---> y=0.5x+2-----> the function represent the graphed line
therefore
the answer is
<span>f(x) = 0.5x + 2
and
</span>y – 1 = 0.5(x + 2)
step 1
find the equation of the line of the graph
let
A (-4,0) B(0,2)
slope m=(y2-y1)/(x2-x1)------> m=(2-0)/(0+4)----> m=1/2----> m=0.5
with m=0.5 and point B (0,2)
y-y1=m*(x-x1)-----> y-2=(0.5)*(x-0)-----> y=0.5x+2
step 2
find the solution
<span>
a) f(x) = 0.2x - 4----> the </span>function not represent the graphed line<span>
b) f(x) = 0.5x + 2----> </span> the function represent the graphed line<span>
c) f(x) = x + 2</span>--> the function not represent the graphed line<span>
d) y – 3 = (x – 2)---> y=x+1---> </span>the function not represent the graphed line<span>
e) y – 1 = 0.5(x + 2)</span>-----> y=0.5x+1+1---> y=0.5x+2-----> the function represent the graphed line
therefore
the answer is
<span>f(x) = 0.5x + 2
and
</span>y – 1 = 0.5(x + 2)
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Answer: (A)
(B) Length varies between 1 and 150
(C) Largest area is 22500ft²
Step-by-step explanation: Suppose length is l and width is w.
The rectangular garden has perimeter of 600ft, which is mathematically represented as
Area of a rectangle is calculated as
Now, we have a system of equations:
Isolate w, so we have l:
w = 300 - l
Substitute in the area equation:
A = l(300 - l)
A = 300l - l²
(A) <u>Function of area in terms of length is given by </u><u>A = 300l - l²</u>
(B) The practical domain for this function is values between 1 and 150.
(C) For the largest area, we need to determine the largest garden possible. For that, we take first derivative of the function:
A' = 300 - 2l
Find the values of l when A'=0:
300 - 2l = 0
2l = 300
l = 150
Replace l in the equation:
w = 300 - 150
w = 150
Now, calculate the largest area:
A = 150*150
A = 22500
<u>The largest area the fence can enclose is </u><u>22500ft².</u>