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3 years ago

Define absolute value

1 answer:

8 0

The absolute value for -5 is 5 on a number line and 5 on a number line for absolute value is 5 if it is a negative number is is the same number but positive if it is a positive number the answer is that number

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What is the y intercept of the line? X=24,36,48 and y= -8, 1, 10

goldenfox [79]

Answer:

y intercept = - 40

Step-by-step explanation:

x 24 36 48

y -8 1 10

y = ax + b

a is the gradient (or slope).

b = y intercept.

a = (48-36) / (10-1) = 12/9

Rewrite to solve for b, which means it starts with b = ...

b = y - 12/9 * x

Substitute one point ( 24 , -8 )

b = -8 - (12/9 * 24)

b = y intercept = - 40

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

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

2 years ago

I need help understanding this. SO please help and give explanation

NikAS [45]

So my understanding from this you have to make it y = mx + b so m stands for the slope there’s different ways to find it as shown in the first picture. I’ll try to make another comment since this only lets me take one picture at a time.