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3 years ago

This problem says 2/3 plus 5/6 but how would you draw the 5/6

Mathematics

2 answers:

kramer3 years ago

6 0

Change 3 to 6 and 2 to 4

Now Do

3/6 + 5/6 = ?

? = 8/6

8/6 = 1 2/6

1 2/6 = 1 1/3

Lesechka [4]3 years ago

3 0

You would need to draw the 5/6 Vertically so that when you combine them it won't look confusing

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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.

8 0

3 years ago

Find the absolute maximum and minimum values of the following function on the given interval. If there are multiple points in a

Bond [772]

$f(x)=3\sin x+3\cos x\implies f'(x)=3\cos x-3\sin x$

$f$ has critical points where $f'=0$ :

$3\cos x-3\sin x=0\implies\cos x=\sin x\implies\tan x=1\implies x=\pm\dfrac\pi4+2n\pi$

where $n$ is any integer. We get solutions in the interval $\left[0,\frac\pi3\right]$ for $n=0$ , for which $x=\frac\pi4$ .

At this critical point, we have $f\left(\frac\pi4\right)=3\sqrt2\approx4.243$ .

At the endpoints of the given interval, we have $f(0)=3$ and $f\left(\frac\pi3\right)=\frac{3+3\sqrt3}2\approx4.098$ .

So we have the extreme values

$\max\limits_{x\in\left[0,\frac\pi3\right]}f=3\sqrt2$

$\min\limits_{x\in\left[0,\frac\pi3\right]}f=3$

8 0

3 years ago

6. Determine how each multiplication or

tangare [24]

A) 27, 81
b) -30,-15
c) -3, 1
d) -320, 1280

3 0

3 years ago

In the figure below, the segments overline RS and overline RT are tangent to the circle centered at Given that OS = 3.6 and RT =

ExtremeBDS [4]

Answer:

RS = 4.8

OT = 3.6

From the Pythagorean Theorem we know:

OR^2 = ST^2 + OT^2

OR^2 = 4.8^2 + 3.6^2

OR ^2 = 23.04 + 12.96

OR^2 = 36

OR = 6

Step-by-step explanation:

8 0

3 years ago

Help me please it’s algebraic expression

Ksju [112]

Answer:

17. xy + z = 51

18 yz - x = 18

21. 5z + ( y - x ) = 20

22. 5x - ( y2 - 4x ) = -10

23 . x2 + y2 - 10z = 70

24. z3 + ( y2 - 4x ) = 1

25. y + xz / 2

26 .3y + x2 / z = 30

Step-by-step explanation:

17.(6 x 8)+3=

48+3=51

18.(8 x 3)- 6 =

24 - 6 = 18

21. 15 + ( 8 - 3 )

15 + 5=

22 .30 - (64 - 24)=

30 - 40=

-10

23. 36 + 64 - 30

100 - 30

24. 9 + (16 - 24 )=

9 + -8 =

25. 8 + 48 / 2

56 / 2

26. 24+ 36/3

60/3=

4 0

3 years ago