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love history [14]
3 years ago
11

A portion of a hiking trail slopes upward at about a 6 angle. To the nearest tenth of a foot, what is the value of x, the distan

ce the hiker traveled along the path, if he has traveled a horizontal distance of 120 feet?
Mathematics
1 answer:
Firdavs [7]3 years ago
4 0
The distance covered by the hiker if he traveled a horizontal distance of 120 ft at and angle of 6° will be given by:
cos θ=adjacent/hypotenuse
let the hypotenuse be h, adjacent =120 ft, θ=6°, thus plugging in the values we shall have:
cos 6=120/h
h=120/cos 6
h=120.661 ft
Answer: 120.661 ft
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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
1 year ago
7. What polynomial remains after the greatest common factor is factored out of the
lara31 [8.8K]

Answer:the answer is A

Step-by-step explanation:

7 0
2 years ago
Solve 2x+ 3=7<br> please ​
blondinia [14]

2x+3=7

2x+3-3=7-3

Whenever moving a number to the other side you have to change the sign .

2x= 4

x=2

7 0
3 years ago
Jonathan’s class has 30 boys. Of the students in his class, 60% are girls. How many girls are in Jonathan’s class?
raketka [301]

Answer:

45 girls

Step-by-step explanation:

  • Remark
  • There are only 2 choices for gender. So if the class is 60% girls, the there must be 100 - 60 = 40% for the boys.
  • But we are told that the boys are 30 in number.
  • Let x = the total number of students.

Solution

  • 40/100 * x = 30              Change the % to a decimal
  • 0.4 x = 30                       Divide both sides by 0.4
  • 0.4 x / 0.4 = 30 / 0.4      Do the division
  • x = 30/0.4
  • x = 75 students in total

Formula

Total of Boys and Girls = Number of boys + number of girls.

Givens

  • Total = 75
  • Number of boys = 30

Solution II

  • 75 = x + 30                 Subtract 30 from each side
  • 75 - 30 = x +30 - 30
  • 45 = x
  • There are 45 girls in the class.
5 0
3 years ago
A tank can be filled by one pipe in four hours and by a second pipe in six hours; and when it is full, the tank can be drained b
just olya [345]
12 hours..........
+ - = 1
Multiply by 12 to clear the fractions
3t + 2t - 4t = 12
t = 12 hrs to fill the pool
3 0
2 years ago
Read 2 more answers
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