Answer:
31.2
Step-by-step explanation:
4(5 + (- 3.5 × - 0.8)
4(5 + 2.8)
4(7.8)
31.2
Gradient is the difference between the y-axis and the x-axis
Gradient = y2-y1/x2-x1
Choose any two point making a triangle
Line B (0,8) (4,0)
This makes a triangle please go look at the graph once again! The the triangle should not have half squares but full!
Line A (5,14)(2,2)
Now plug in the variable
The difference between the Y and X
Line B 0-8/4-0
Line B gradient =-2
Line A 2-14/2-5
Line A gradient = +4
Finalized answer
Line B gradient = -2
Line A gradient = +4
Answer:
Any equation of the line that is different from y=x is the solution to the problem.
Step-by-step explanation:
step 1
Find the slope of the line that passes through the points (1, 1) and (5, 5)
m=(5-1)/(5-1)=1
step 2
Find the equation of the line into slope point form
y-y1=m(x-x1)
we have m=1
(x1,y1)=(1,1)
substitute
y-1=(1)(x-1)
y=x-1+1
y=x
therefore
Any equation of the line that is different from y=x is the solution to the problem.
The dimensions of the enclosure that is most economical to construct are; x = 14.22 ft and y = 22.5 ft
<h3>How to maximize area?</h3>
Let the length of the rectangular area be x feet
Let the width of the area = y feet
Area of the rectangle = xy square feet
Or xy = 320 square feet
y = 320/x -----(1)
Cost to fence the three sides = $6 per foot
Therefore cost to fence one length and two width of the rectangular area
= 6(x + 2y)
Similarly cost to fence the fourth side = $13 per foot
So, the cost of the remaining length = 13x
Total cost to fence = 6(x + 2y) + 13x
Cost (C) = 6(x + 2y) + 13x
C = 6x + 12y + 13x
C = 19x + 12y
From equation (1),
C = 19x + 12(320/x)
C' = 19 - 3840/x²
At C' = 0, we have;
19 - 3840/x² = 0
19 = 3840/x²
19x² = 3840
x² = 3840/19
x = √(3840/19)
x = 14.22 ft
y = 320/14.22
y = 22.5 ft
Read more about Maximization of Area at; brainly.com/question/13869651
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QPN and TSU !!!! they are both angles on the exterior with a larger value!!
