Answer:
Micheal sent 40 messages, Christine sent 20, and Dave sent 11.
Step-by-step explanation:
Christine, Dale, and Michael sent a total of 71 messages
C + D + M = 71
Dale sent 9 fewer messages than Christine
D = C - 9
Michael sent 2 times as many messages as Christine
M = 2C
Plug-in the numbers.
C + C - 9 + 2C = 71
4C - 9 = 71
4C = 80
C = 20
Now, plug in to other equations for other results.
D = (20) - 9
D = 11
M = 2(20)
M = 40
Micheal sent 40 messages, Christine sent 20, and Dave sent 11.
Verify?
40 + 20 + 11 = 71.
The last option: In his recursive formula, he should have the term an−1 instead of −4 to get an=an−1−3.
Answer:
17 1/2 in²
Step-by-step explanation:
a=l×w
or in this case a=h×b
2 1/2×7= 17 1/2 in²
Your question is:
![\frac{8u}{u^2-18u+81} -\frac{72}{u^2-18u+81}](https://tex.z-dn.net/?f=%5Cfrac%7B8u%7D%7Bu%5E2-18u%2B81%7D%20-%5Cfrac%7B72%7D%7Bu%5E2-18u%2B81%7D)
Then, the above formula equate:
![\frac{8u-72}{u^2-18u+81}](https://tex.z-dn.net/?f=%5Cfrac%7B8u-72%7D%7Bu%5E2-18u%2B81%7D)
Factor out 8 from the numerator we get:
![\frac{u-9}{u^2-18u+81}](https://tex.z-dn.net/?f=%5Cfrac%7Bu-9%7D%7Bu%5E2-18u%2B81%7D)
Factorization of the denominator:
![\frac{u-9}{(u-9)^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bu-9%7D%7B%28u-9%29%5E2%7D)
Simplifying the numerator and the denominator by (u-9) we get:
Answer:
Area = 2500 square feet is the largest area enclosed
Step-by-step explanation:
A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fencing
Let x be the length of each box and y be the width of the box
Perimeter of the box= 3(length ) + 4(width)
![200=3x+4y](https://tex.z-dn.net/?f=200%3D3x%2B4y)
solve for y
![200=3x+4y](https://tex.z-dn.net/?f=200%3D3x%2B4y)
![200-3x=4y](https://tex.z-dn.net/?f=200-3x%3D4y)
divide both sides by 4
![y=50-\frac{3x}{4}](https://tex.z-dn.net/?f=y%3D50-%5Cfrac%7B3x%7D%7B4%7D)
Area of the rectangle = length times width
![Area = 3x \cdot y](https://tex.z-dn.net/?f=Area%20%3D%203x%20%5Ccdot%20y)
![Area = 3x \cdot (50-\frac{3x}{4})](https://tex.z-dn.net/?f=Area%20%3D%203x%20%5Ccdot%20%2850-%5Cfrac%7B3x%7D%7B4%7D%29)
![A=150x-\frac{9x^2}{4}](https://tex.z-dn.net/?f=A%3D150x-%5Cfrac%7B9x%5E2%7D%7B4%7D)
Now take derivative
![A'=150-\frac{9x}{2}](https://tex.z-dn.net/?f=A%27%3D150-%5Cfrac%7B9x%7D%7B2%7D)
Set it =0 and solve for x
![0=150-\frac{9x}{2}](https://tex.z-dn.net/?f=0%3D150-%5Cfrac%7B9x%7D%7B2%7D)
![150=\frac{9x}{2}](https://tex.z-dn.net/?f=150%3D%5Cfrac%7B9x%7D%7B2%7D)
multiply both sides by 2/9
![x=\frac{100}{3}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B100%7D%7B3%7D)
![A''=-\frac{9}{2}](https://tex.z-dn.net/?f=A%27%27%3D-%5Cfrac%7B9%7D%7B2%7D)
For any value of x, second derivative is negative
So maximum at x= 100/3
, replace the value of x
![A=150(\frac{100}{3})-\frac{9(\frac{100}{3})^2}{4})](https://tex.z-dn.net/?f=A%3D150%28%5Cfrac%7B100%7D%7B3%7D%29-%5Cfrac%7B9%28%5Cfrac%7B100%7D%7B3%7D%29%5E2%7D%7B4%7D%29)
Area = 2500 square feet is the largest area enclosed