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3 years ago

Determine the y-intercept and slope and then write the equation for this table.*

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

4 0

Answer:

The y-intercept is 15 and the slope is 5.5

y=5.5x+15

Step-by-step explanation:

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What is the first step to solve the following system of equations using substitution?

Akimi4 [234]

Answer:

Step 1: Solve one of the equations for one of the variables. Let's solve the first equation for x

Thus, option A) is true.

The solution to the system of equations be:

$x = 4, y = 5$

Step-by-step explanation:

It is important to remember that when we solve the system of equations, the first step we need to do is to solve one of the equations for one of the variables.

Given the system of equations

$x - y = -1$

$2x - y = 3$

Step 1: Solve one of the equations for one of the variables. Let's solve the first equation for x

$x-y=-1$

Add y to both sides

$x-y+y=-1+y$

$x=-1+y$

Thus, option A) is true.

NOW LET US SOLVE THE REMAINING PORTION

$\mathrm{Subsititute\:}x=-1+y$ to solve for y

$\begin{bmatrix}2\left(-1+y\right)-y=3\end{bmatrix}$

$-2+y=3$

$y=5$

For x = -1 + y

substitute y = 5

$x=-1+5$

$x=4$

Thus, the solution to the system of equations be:

$x = 4, y = 5$

3 0

2 years ago

If a and b are real numbers, what property of real numbers is illustrated by the equation below? 4a + 4b = 4(a + b)

Tresset [83]

D. Distributed Property of Multiplication over Addition. Happy Thanksgiving. Hope this helps.

7 0

3 years ago

(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$