The interval where the function is nonlinear and decreasing is 0 < x < 4
<h3>How to determine the interval where the function is nonlinear and decreasing?</h3>
The straight lines on the graph are the intervals where the graph is linear
This means that the straight lines on the graph will not be considered
Considering the curve, the graph decrease from x = 0 to x = 4
This can be rewritten as:
0 < x < 4
Hence, the interval where the function is nonlinear and decreasing is 0 < x < 4
Read more about function intervals at:
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Answer:
-an - 4a + 3n + 12
Step-by-step explanation:
(-a + 3)(n + 4)
= -an - 4a + 3n + 12
Part A: 216 students/ 12 classrooms= 18 students: 1 classroom.
Part B: 104 students/ 4 classrooms= 26 students: 1 classroom.
The average ratio of students to classrooms in both school is:
[(18 students: 1 classroom)+ (26 students: 1 classroom)] /2= 22 students: 1 classroom.
26 students - 22 students= 4 students.
There are 4 students who would have to transfer from School B to School A for the ratios of students to classrooms at both schools to be the <span>same, which is 22 students: 1 classroom.
Hope this helps~</span>
Answer:
The answer is "
"
Step-by-step explanation:
The rectangle should also be symmetrical to it because of the symmetry to the y-axis The pole of the y-axis. Its lower two vertices are (-x,0). it means that
and (-x,0), and (x,0). Therefore the base measurement of the rectangle is 2x. The top vertices on the parabola are as follows:
The calculation of the height of the rectangle also is clearly 16-x^2, (-x,16,-x^2) and (x,16,-x^2).
The area of the rectangle:
![A(x)=(2x)(16-x^2)\\\\A(x)=32x-2x^3](https://tex.z-dn.net/?f=A%28x%29%3D%282x%29%2816-x%5E2%29%5C%5C%5C%5CA%28x%29%3D32x-2x%5E3)
The local extremes of this function are where the first derivative is 0:
![A'(x)=32-6x^2\\\\32-6x^2=0\\\\x= \pm\sqrt{\frac{32}{6}}\\\\x= \pm\frac{4\sqrt{3}}{3}\\\\](https://tex.z-dn.net/?f=A%27%28x%29%3D32-6x%5E2%5C%5C%5C%5C32-6x%5E2%3D0%5C%5C%5C%5Cx%3D%20%5Cpm%5Csqrt%7B%5Cfrac%7B32%7D%7B6%7D%7D%5C%5C%5C%5Cx%3D%20%5Cpm%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%5C%5C%5C%5C)
Simply ignore the negative root because we need a positive length calculation
It wants a maximum, this we want to see if the second derivative's profit at the end is negative.
![A''\frac{4\sqrt{3}}{3} = -12\frac{4\sqrt{3}}{3}](https://tex.z-dn.net/?f=A%27%27%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%20%3D%20-12%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%3C0%5C%5C%5C%5C2.%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%3D%20%5Cfrac%7B8%5Csqrt%7B3%7D%7D%7B3%7D%5C%5C%5C%5C%5Cvertical%20%5C%20dimension%5C%5C%5C%5C16-%28%5Cfrac%7B4%5Csqrt%7B3%7D%7D%7B3%7D%29%5E2%3D%20%5Cfrac%7B32%7D%7B3%7D)