HELP PLEASE!!!

Mathematics

1 answer:

VMariaS [17]3 years ago

7 0

Answer:

Neither

Step-by-step explanation:

First, you need to put your equations in y=mx+b form

Your first equation...

5x + 4y = 3

4y = 3-5x

y = -5/4x + 3/4

Your second equation...

5x-4y = -3

-4y = -5x -3

y = 5/4x + 3/4

The slopes aren't the same and are not reciprocals, so the answer would be neither.

You might be interested in

0.5 as a repeating decimal to a fraction

nika2105 [10]

Those are just repeating equivalent decimals

5 0

3 years ago

Hey I Need Help With This. What Is the answer?<br>

scoundrel [369]

Answer:

22%

Step-by-step explanation:

out of 100, 22 boxes are shaded

3 0

3 years ago

Using Euler’s Method with a step size of h=0.5, estimate the value for y(3) for the differential equation xy dy/dx= 1+ln(x^2), w

lora16 [44]

Euler's method uses the recurrence relation

$y_{n+1}=y_n+hf(x_n,y_n)$

to approximate the value of the solution $y(x)$ to the ODE $y'=f(x,y)$ .

$xy\dfrac{\mathrm dy}{\mathrm dx}=1+\ln(x^2)\implies y'=\dfrac{1+\ln(x^2)}{xy}=f(x,y)$

With a step size of $h=0.5$ , there will only be two steps necessary to find the approximate value of $y(3)$ based on the initial point $y(2)=5$ . See the attached table below for the computation results.

To demonstrate how the table is generated: Since $y(2)=5$ , you are using $(x_0,y_0)=(2,5)$ .

$y_1=y_0+hf(x_0,y_0)$
$y_1=5+0.5\times\dfrac{1+\ln(2^2)}{2\times5}$
$y_1\approx5.1193$

The next point then uses $x_1=x_0+0.5=2.5$

$y_2=y_1+hf(x_1,y_1)$
$y_1=y_1+0.5\times\dfrac{1+\ln(2.5^2)}{2.5y_1}$
$y_2\approx5.230$