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2 years ago

9

In order to get a good workout on your treadmill, you figure you have to run 528 feet in one minute. What are the possible numbe

rs of miles you have to run in 20 minutes to exceed this rate? (Hint: 1 mile = 5280 feet)
PLEASE PROVIDE AN EXPLANATION, thank you!

1 answer:

Scrat [10]2 years ago

5 0

Answer:

2 miles.

Step-by-step explanation: Because 528 times 20 is 10560 then divide by 5280 equaling 2 miles.

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Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

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Help me on this question! Jake can carry 6 1/4 pounds of wood in from the barn.His father can carry 1 5/7 times as much as jake.

ioda

$\text{ Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }$

<em><u>Solution:</u></em>

From given question,

Number of pounds Jake carry = $6\frac{1}{4} \text{ pounds }$

Number of pounds his father carry is $1\frac{5}{7}$ times as much as jake

To find: Number of pounds Jake father can carry

<em><u>Convert the mixed fractions to improper fractions</u></em>

Multiply the whole number part by the fraction's denominator.

Add that to the numerator.

Then write the result on top of the denominator

$\rightarrow 6\frac{1}{4} = \frac{4 \times 6+1}{4} = \frac{25}{4}\\\\\rightarrow 1\frac{5}{7} = \frac{7 \times 1+5}{7} = \frac{12}{7}$

<em><u>Then according to question,</u></em>

$\text{Pounds father carry } = \frac{12}{7} \text{ times the pounds jake carry }\\\\\text{Pounds father carry } = \frac{12}{7} \times \frac{25}{4}\\\\\text{Pounds father carry } = \frac{75}{7}\\\\\text{In terms of mixed fractions, }\\\\\text{Pounds father carry } = \frac{75}{7} = 10\frac{5}{7}$

$\text{ Thus, Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }$

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