All categories

2 years ago

What are the solutions to the system of the equations graphed below?

Mathematics

2 answers:

xz_007 [3.2K]2 years ago

8 0

A (2,0) and (0,-8) because it’s shown in the picture that’s the red line indicates the place it needs to be in

Zigmanuir [339]2 years ago

4 0

Answer:

Step-by-step explanation:

Where the graphs intersect each other is where the solution is. The graphs intersect each other at the points (0,-8) and (4,8). Hope this helps!

You might be interested in

Solve for x in the equation x^2-10x+25=35.

KatRina [158]

So, this is too vague but I'll solve for both quadratic formula and find the discriminant.

Quadratic formula: x=5+√35 OR x=5−√35
Finding the discriminant: 141

3 0

3 years ago

Read 2 more answers

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

What is the product of -1/2 x 1/8 in simplest form

hjlf

Answer: Edit: -0.0625

Step-by-step explanation:

3 0

3 years ago

How do I solve this?

bogdanovich [222]

Answer:

Step-by-step explanation:

$5(x+7)^4=\boxed5\cdot\boxed{(x+7)}\cdot\boxed{(x+7)}\cdot(x+7)\cdot(x+7)\\\\35(x+7)^2=\boxed5\cdot7\cdot\boxed{(x+7)}\cdot\boxed{(x+7)}\\\\15(x+7)^3=\boxed5\cdot3\cdot\boxed{(x+7)}\cdot\boxed{(x+7)}\cdot(x+7)\\\\GCF\bigg(5(x+7)^4,\ 35(x+t)^2,\ 15(x+7)^3\bigg)=\boxed5\cdot\boxed{(x+7)}\cdot\boxed{(x+7)}\\\\=5(x+7)^2$

5 0

3 years ago

Please someone help me answer this!!

jekas [21]

Answer:

a^10

Step-by-step explanation:

5*2=10

3 0

3 years ago