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weeeeeb [17]
4 years ago
14

Toni can carry up to 18 lb in her backpack. Her lunch weighs 1 lb, her gym clothes weigh 2 lb, and her books (b) weigh 3 lb each

. How many books can she carry in her backpack?
Mathematics
1 answer:
MrRissso [65]4 years ago
3 0
<span>Toni can carry five books in her backpack. You must start by subtracting the weight of the other required items. Toni's lunch which is one pound and gym clothes which are two pounds. This can be calculated by 18-1-2=15. Toni has 15 pounds left to carry books. She should divide this by the weight of each book, which is three pounds. 15/3=5. Therefore, Toni can carry five books.</span>
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A competitive diver dives from a 33-foot high diving board. The height of the diver in feet after 't' seconds is given by u(t) =
ludmilkaskok [199]

Answer:

t  = 0.375s

Step-by-step explanation:

Given

h(t) = -16t^2 + 4t + 33 --- driver 1

Rate = 2ft/s -- driver 2

height = 33ft

Required

The time they passed each other

First, we determine the function of driver 2.

We have that:

Rate = 2ft/s and height = 33ft

So, the function is:

h_2(t) = Height - Rate * t

h_2(t) = 33 - 2t

The time they drive pass each other is calculated as:

h(t) = h_2(t)

-16t^2 + 4t + 33= 33 - 2t

Collect like terms

-16t^2 + 4t + 2t= 33 - 33

-16t^2 + 6t= 0

Divide through by 2t

-8t + 3= 0

Solve for -8t

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Solve for t

t  = \frac{-3}{-8}

t  = 0.375s

7 0
3 years ago
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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