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3 years ago

How do you solve each system by graphing

Mathematics

1 answer:

abruzzese [7]3 years ago

8 0

See the attached graph

<h2>Explanation:</h2>

Hello! Remember to write complete questions in order to get good and exact answers. Here you don't provide any system so I couldn't help you in an exact way, but in a general manner. When solving systems by graphing we just need to find the points of intersection of the graphs of the functions that built up the system.

Suppose you have the following system of linear equation:

$\left\{ \begin{array}{c}y=2x+3\\3y=8x+27\end{array}\right.$

yBy using graphing tools, we get the graph shown below. As you can see, the point of intersection of both lines is $(-9,-15)$

<h2>Learn more:</h2>

Solving systems by graphing: brainly.com/question/13799715

#LearnWithBrainly

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X+4Y=-8 find the x intercept: find the y intercept: graph the line.

galben [10]

Answer:

x-intercept = (-8, 0)

y-intercept = (0, -2)

Step-by-step explanation:

Given linear equation is: x + 4y = - 8

For x - intercept: Plug y = 0 in the above equation.

$\implies x +4(0) = -8\\\\\implies x +0 =-8\\\\ \implies x = -8\\\\ \boxed{\red{\bold{\implies x- intercept = (-8,\:0)}}}$

For y - intercept: Plug x = 0 in the given linear equation.

$\implies 0 +4y = -8\\\\\implies 4y=-8\\\\ \implies y = \frac{-8}{4}\\\\ \implies y = -2\\\\ \boxed{\purple{\bold{\implies y- intercept = (0,\:-2)}}}$

7 0

2 years ago

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

y''(t) + p(t) y'(t) + q(t) y(t) = f(t)

so the first thing you need to do is divide both sides of your equation by t :

y'' + (2t - 1)/t y' - 2/t y = 7t

You're looking for a solution of the form

$y=y_1u_1+y_2u_2$

where

$u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

$u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt$

and W denotes the Wronskian determinant.

Compute the Wronskian:

$W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}$

Then

$u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t$

$u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}$

The general solution to the ODE is

$y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}$

which simplifies somewhat to

$\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}$

4 0

2 years ago

BRAINLIEST ASAP! PLEASE ANSWER!!!!!

hodyreva [135]

Similarities: They both are polynomials of degree 2, both of their graphs is a parabola, both have either 2 or 0 real solutions, they are both continuous functions over R
(DOS= difference of two squares, PST=perfect square trinomial
Differences: PST has three terms, whereas the difference of squares has 2. PST's factors are both the same, whereas DOS's elements are conjugates of each other. DOS can always be factored into two distinct polynomials with rational coefficients, whereas PST has two same polynomial factors.

7 0

3 years ago

47. What is the 12th term of the geometric sequence?

GrogVix [38]

Answer:

Step-by-step explanation:

The rule is 2x.

So 3,6,12,24,48,96,192,384,768,1536,3072,6144,12288,....

The question asked for the 12th term so 6144.

5 0

2 years ago

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