All categories

2 years ago

Which graph represents the solution to the compound inequality?

Mathematics

1 answer:

alisha [4.7K]2 years ago

7 0

Answer:

We know that the slope or rate of change of the function can be:

positive

negative

zero, or

undefined

Function 1

From the function 1 graph, it is clear that the graph is a horizontal line. We must note that the horizontal line has a slope or rate of change zero. The reason is that the horizontal line can not rise vertically. i.e. y₂-y₁=0

so using the slope formula

Rate of change = m = y₂-y₁ / x₂-x₁

Taking two points (x₁, y₁) = (0, 4), (x₂, x₁) = (1, 4)

Rate of change = m = 4-4 / 1-0

Rate of change = m = 0/1

Rate of change = m = 0

Thus, the rate of change of function 1 is zero.

Function 2

We know the slope-intercept form of linear equation is

where m is the rate of change or slope of the function and b is the y-intercept

Given the function

comparing with the slope-intercept form i.e. y = mx+b

Therefore, the rate of change of function 2 = m = 8

Conclusion

The rate of change of function 1 = 0

The rate of change of function 2 = 8

8 - 0 = 8

Therefore, function 2 has 8 more rate of change of function than the rate of change of function 1.

You might be interested in

How many solutions does 3 (y-2) = 3y - 6 have

Marrrta [24]

Answer:

Infinitely many solutions

Step-by-step explanation:

Simplify this

3 (y-2) = 3y-6

3 (y) + 3 ( -2) = 3y-6

3y-6=3y-6

Because these two equations are equal to each other, it is infinitely many solutions.

If this helps please mark as brainliest

3 0

3 years ago

FIND THE UNIT RATE!!!!

uranmaximum [27]

Answer:

It is 120 calories per serving.

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Ellen bought soccer shorts on sale for $6 off the regular price of $40. what percent did she save?

jeyben [28]

Ellen bought soccer shorts on sale for $6 off the regular price of $40. What Percent did she save?

The answer is 15%

7 0

4 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 :

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