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3 years ago

An office manager spent $4870 on a

Mathematics

1 answer:

Andrews [41]3 years ago

7 0

Answer:

6 chairs

8 desks

Step-by-step explanation:

First, we assume that all 14 items he bought were chairs, if that was to be true then the office manager would spend...

125 x 14 = 1750$

We see that he spent less then he should have so now we start to substitute desks instead of chairs, and we have to keep in mind that if we change 1 chair to a desk then the total will increase by...

515 - 125 = 390$

Now we just find the difference between the needed total and the total that we get if the office manager would by only chairs and then divide the result by 390 in order to find the number of desks, and so we get...

(4870 - 1750) / 390 = 3120 / 390 = 8 desks

And now that we know the number of desks we just subtract the number of desks from 14 and get the number of chairs, and so...

14 - 8 = 6 chairs

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

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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pychu [463]

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Tems11 [23]

Answer:

Step-by-step explanation:

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ololo11 [35]

-- The graph looks like a line that passes through the origin,
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-- Point #1 on the line:
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. . . . . Call the first one 'x'. Call the second one 'y'.

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. . . . . Write it down twice.
. . . . . Call the first one 'x'. Call the second one 'y'.

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Rinse and repeat, as many times as you like,
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Sveta_85 [38]

Answer:

question 9

= –12

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Step-by-step explanation:

question 9=

u= –16+4= –12

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3 years ago