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DerKrebs [107]
2 years ago
13

Where is mean by g(x)​

Mathematics
1 answer:
matrenka [14]2 years ago
5 0

Answer:

g(x)=2x or h(x) =2x,,,, mean the same thing

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Simplify 2m - [n - (m - 2n)].<br><br> -3m - n<br> 3m - n<br> -3m - 3n<br> 3m - 3n
VMariaS [17]
2m - [n - (m - 2n)]

2m - n + (m - 2n)

2m - n + m - 2n

2m + m - n - 2n

3m - 3n

Answer is D.
5 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.

8 0
3 years ago
Suppose the area of a circle is 49π. What is its radius
pentagon [3]
The radius is 7 what it is telling you is that you already did the square 4^2*3.14=49 pi. 
4 0
3 years ago
The number of students enrolled at a college is 17,000 and grows 6% each year.
Ad libitum [116K]

Answer:

1020 new students per year.

Step-by-step explanation:

5 0
3 years ago
What is the equation of the line shown in the graph?
postnew [5]

The slope-intercept form:

y=mx+b

m - slope

b - y-intercept

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-3, 1) and (2, -4). Substitute:

m=\dfrac{-4-1}{2-(-3)}=\dfrac{-5}{5}=-1

Therefore we have the equation of a line

y=-1x+b\to y=-x+b

Put the coordinates of the point (-3, 1) to the equation:

1=-(-3)+b

1=3+b      <em>subtract 3 from both sides</em>

-2=b\to b=-2

<h3>Answer: y = -x - 2</h3>
4 0
3 years ago
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