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3 years ago

When solving 20x+5y=40 for y

Mathematics

2 answers:

Aleks04 [339]3 years ago

6 0

The answer is y=-4x+8. The paper shows how to solve the equation step by step.

Paha777 [63]3 years ago

3 0

Simplifying
20x + -5y = 40

Solving
20x + -5y = 40

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '5y' to each side of the equation.
20x + -5y + 5y = 40 + 5y

Combine like terms: -5y + 5y = 0
20x + 0 = 40 + 5y
20x = 40 + 5y

Divide each side by '20'.
x = 2 + 0.25y

Simplifying
x = 2 + 0.25y

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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

Nathan cut a square tile in half for his kitchen floor design. He made one cut along a diagonal from one vertex to another verte

Amiraneli [1.4K]

<span>3. Two isosceles right triangles

Every square has 4 right angles, so each triangle will have one right triangle, which makes the triangles, right triangles.

They are isosceles because all sides of the square are the same, though the diagonal is not.

Hope this helps :)

</span>

4 0

3 years ago

The distance d, in centimeters, that a diving board bends below its resting position when you stand at its end is dependent on y

kaheart [24]

Answer:

d(1) = -3 and d(2)= -28

Step-by-step explanation:

Given that the distance d, in centimeters, that a diving board bends below its resting position when you stand at its end is dependent on your distance x, in meters, from the stabilized point

$d(x)=-4x^3+x^2$