A golfer wants to determine if the type of driver she uses each year can be used to predict the amount of improvement in her gam
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Step-by-step explanation:
We have been given an equation y+6=45(x+3) in point slope form.
It says to use the point and slope from given equation to create the graph.
So compare equation y+6=45(x+3) with point slope formula
y-y1=m(x-x1)
we see that m=45, x1=-3 and y1=-6
Hence first point is at (-3,-6)
slope m=45 is positive so to find another point, previous point will move 45 units up then 1 unit right and reach at the location (-2,39).
Now we just graph both points (-3,-6) and (-2,39) and join them by a straight line. Final graph will look like the attached graph.
This is a problem of maxima and minima using derivative.
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:
We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =
Surface area of the rectangular sides =
Therefore, the total area of the cube is:
Isolating the variable y in terms of x:
Substituting this value in V:
Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:
Solving for x:
Solving for y:
Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft
And its volume is:
In the figure shown below we have the representation of this problem, so we know that the base of this bin is square. We also know that there are four square rectangles sides. This bin is a cube, therefore the volume is:
V = length x width x height
That is:
We also know that the <span>bin is constructed from 48 square feet of sheet metal, s</span>o:
Surface area of the square base =
Surface area of the rectangular sides =
Therefore, the total area of the cube is:
Isolating the variable y in terms of x:
Substituting this value in V:
Getting the derivative and finding the maxima. This happens when the derivative is equal to zero:
Solving for x:
Solving for y:
Then, <span>the dimensions of the largest volume of such a bin is:
</span>
Length = 4 ft
Width = 4 ft
Height = 2 ft
And its volume is: