The first step for solving this problem is to multiply both sides of the bottom equation by -3.
Add the two equations together.
6x - 9y - 6x + 9y = 16 - 21
Eliminate the opposites.
-9y + 9y = 16 - 21
Remember that the sum of two opposites equals 0,, so the equation becomes the following:
0 = 16 - 21
Calculate the difference on the right side of the equation.
0 = -5
This means that the statement
is false for any value of x and y. That means that the answer to your question is (x,y) ∈ ∅,, or no solution.
Let me know if you have any further questions.
:)
Answer:
1
Step-by-step explanation:
Uhh my brains hurts looking at that
<h3>
Answer: 140625 </h3>
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Explanation:
Replace every copy of "w" with "x"
So the function we graph is y = (1500-2x)*x/2
I used GeoGebra to make the graph below. Note the scale on the xy axis. In this case, x is incrementing by 100 while y is incrementing by 50,000. The scale is important so you have a good viewing window. You don't have to have this scale exactly, but something close to it should do the trick. Without the proper scale, you probably won't see the curve at all. You may see a straight line instead, or it may appear completely blank.
You can use Desmos to make the graph. Just keep in mind the scale of course.
Once you have the graph set up, use the max feature of your graphing calculator or graphing software to locate the vertex point. In GeoGebra, the function I used is "Max". I typed in Max(f, 0, 800) where f is the function mentioned earlier. With Desmos, you simply need to click on the curve itself to have the max point show up. Click on the vertex point to have the coordinates listed.
The max point is located at A = (375, 140625) as shown in the diagram below. The x coordinate is the value of "w" that we replaced earlier. So a width of w = 375 feet corresponds to the max area of 140625 square feet.
Side note: later on in your math career, you have the option to use calculus to solve a problem like this. Though for now, we can rely on a graphing calculator to get the job done quickly.
