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3 years ago

Which equation can pair with x-y=-2 to create a consistent and dependent system?

Mathematics

1 answer:

MrRissso [65]3 years ago

3 0

See the explanation

<h2>Explanation:</h2>

A system that has one or infinitely many solutions is called consistent. If an equation in a system tells us no new information then the equations of the system are dependent. In other words, to find an equation that creates a consistent and dependent system with the given equation we have to get the same line:

The given line is:

$x-y=-2$

If we multiply both sides of the equation by a constant we will have the same line when plotting, therefore let's multiply by 3:

$3(x-y)=3(-2) \\ \\ 3x-3y=-6$

So a system of two linear equation that is consistent and dependent is:

$\left \{ {{x-y=-2} \atop {3x-3y=-6}} \right.$

<h2>Learn more:</h2>

Graph of lines: brainly.com/question/14434483#

#LearnWithBrainly

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What fraction is equal to 0.7

olga nikolaevna [1]

The equal fraction is 7/10

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3 years ago

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(10) Both red line and blue line trains left the station at 5:00 pm. If both

scoundrel [369]

Answer:

5:40

Step-by-step explanation:

This is a problem involving the least common difference.

If you know that the red and blue trains left at the same time at 5, you know that another red train will leave at 5:08. Another blue train at 5:10.

The way to solve this will be to write out the factors of 8 and 10 and find the smallest number that they overlap.

Red:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80

Blue:

10, 20, 30, 40

You see that after 40 mnutes, they are both leaving the station again. After 40 minutes, at 5:40, they are both leaving.

6 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 :

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

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

5 0

3 years ago

What is the factored form of x^2 - 196 ?

Luba_88 [7]

D. (x-14)(x+14)
a negative and a positive multiplied makes a negative outcome so -14•+14=-196. x•x is = to x^2

8 0

3 years ago

Read 2 more answers

The tensile strength of a metal part is normally distributed with mean 40 pounds and standard deviation 5 pounds. If 50,000 part

nalin [4]

Answer:

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

7933 parts

b) How many would have a tensile strength in excess of 48 pounds?

2739.95 parts

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

a) how many would you expect to fail to meet a minimum specification limit of 35-pounds tensile strength?

z = (x-μ)/σ

x = 35 μ = 40 , σ = 5

z = 35 - 40/5

= -5/-5

= -1

Determining the Probability value from Z-Table:

P(x<35) = 0.15866

Converting to percentage = 15.866%

We are asked how many will fail to meet this specification

We have 50,000 parts

Hence,

15.866% of 50,000 parts will fail to meet the specification

= 15.866% of 50,000

= 7933 parts

Therefore, 7933 parts will fail to meet the specifications.

b) How many would have a tensile strength in excess of 48 pounds?

z = (x-μ)/σ

x = 48 μ = 40 , σ = 5

z = 48 - 40/5

z = 8/5

z = 1.6

P-value from Z-Table:

P(x<48) = 0.9452

P(x>48) = 1 - P(x<48)

1 - 0.9452

= 0.054799

Converting to percentage

= 5.4799%

Therefore, 5.4799% will have an excess of (or will be greater than) 48 pounds

We are asked, how many would have a tensile strength in excess of 48 pounds?

This would be 5.4799% of 50,000 parts

= 5.4799% × 50,000

= 2739.95

Therefore, 2739.95 parts will have a tensile strength excess of 48 pounds

4 0

3 years ago