Do the parentheses first :) ʕ•ᴥ•ʔ
None of the above? If the width is 25 feet long, the length would have to be 20 feet, which is shorter than the width.
Do 6x1=6. 6 is in the question and is also the answer.
See the picture attached to better understand the problem
Let
x----------> The distance from Chester to Durbin
we know that
![10 \frac{3}{5} = \frac{(10*5+3)}{5} \\ \\ = \frac{53}{5} miles](https://tex.z-dn.net/?f=10%20%20%5Cfrac%7B3%7D%7B5%7D%20%3D%20%5Cfrac%7B%2810%2A5%2B3%29%7D%7B5%7D%20%20%5C%5C%20%5C%5C%20%20%3D%20%5Cfrac%7B53%7D%7B5%7D%20miles)
![12 \frac{1}{2} = \frac{(12*2+1)}{2} \\ \\ = \frac{25}{2} miles](https://tex.z-dn.net/?f=12%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%5Cfrac%7B%2812%2A2%2B1%29%7D%7B2%7D%20%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B25%7D%7B2%7D%20miles%20)
![10 \frac{3}{5} +12 \frac{1}{2} +x=35](https://tex.z-dn.net/?f=10%20%20%5Cfrac%7B3%7D%7B5%7D%20%2B12%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2Bx%3D35)
subtitute
![\frac{53}{5} + \frac{25}{2} +x=35](https://tex.z-dn.net/?f=%20%5Cfrac%7B53%7D%7B5%7D%20%2B%20%5Cfrac%7B25%7D%7B2%7D%20%2Bx%3D35)
solve for x
![x=35-( \frac{53}{5}+ \frac{25}{2} ) \\ \\ x=35- \frac{(2*53+5*25)}{10} \\ \\ x=35- \frac{231}{10} \\ \\ x=35-23.10 \\ \\ x=11.9](https://tex.z-dn.net/?f=x%3D35-%28%20%5Cfrac%7B53%7D%7B5%7D%2B%20%5Cfrac%7B25%7D%7B2%7D%20%29%20%5C%5C%20%5C%5C%20x%3D35-%20%5Cfrac%7B%282%2A53%2B5%2A25%29%7D%7B10%7D%20%5C%5C%20%5C%5C%20x%3D35-%20%5Cfrac%7B231%7D%7B10%7D%20%5C%5C%20%5C%5C%20x%3D35-23.10%20%5C%5C%20%5C%5C%20x%3D11.9%20)
![x=11 \frac{9}{10}](https://tex.z-dn.net/?f=x%3D11%20%5Cfrac%7B9%7D%7B10%7D)
miles
the answer is
miles
Answer:
a. 0.588
b. 0.0722
c. 4.576 sqft/sec
Step-by-step explanation:
Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem,
because it is a right angle triangle.
It is given that ![\frac{db}{dt} = 1](https://tex.z-dn.net/?f=%20%5Cfrac%7Bdb%7D%7Bdt%7D%20%3D%201%20)
Now differentiate (1) with respect to t (time) :
![\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B2h%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%202b%5Cfrac%7Bdb%7D%7Bdt%7D%20%3D%200%20%5Cimplies%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%3D%20-%5Cfrac%7Bb%7D%7Bh%7D%20%5Cfrac%7Bdb%7D%7Bdt%7D%7D)
![\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%3D-%5Cfrac%7Bb%7D%7B%5Csqrt%7B256%20-%20b%5E2%7D%7D%20%5Cfrac%7Bdb%7D%7Bdt%7D%20%3D%20-%5Cfrac%7B8%7D%7B13.856%7D%20%5Ctimes%201%20%3D%20-0.588%7D)
The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.
b. From the diagram, infer that
. When b = 8, then
.
Differentiate the above equation w.r.t t
![\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B16%20%5Ccos%7B%5Ctheta%7D%20%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bdb%7D%7Bdt%7D%20%5Cimplies%20%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%20%3D%20%5Cfrac%7B1%7D%7B16%20%5Ccos%7B%5Ctheta%7D%7D%20%3D%20%5Cfrac%7B1%7D%7B13.856%7D%20%3D%20%5Cmathbf%7B0.0722%7D%7D)
c. The area of the triangle is given by
. Differentiating w.r.t t,
![\displatstyle{\frac{dA}{dt} = 0.5 b \frac{dh}{dt} + 0.5 h \frac{db}{dt}}](https://tex.z-dn.net/?f=%5Cdisplatstyle%7B%5Cfrac%7BdA%7D%7Bdt%7D%20%3D%200.5%20b%20%5Cfrac%7Bdh%7D%7Bdt%7D%20%2B%200.5%20h%20%5Cfrac%7Bdb%7D%7Bdt%7D%7D)
Plugging in b = 8, h = 13.856,
,