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2 years ago

1. Find the slope of the line:

Mathematics

1 answer:

Flura [38]2 years ago

8 0

Answer:

Step-by-step explanation:

Slope=rise/run

=4 blocks/ 1 block

Since 1 block=5 units,

slope= $\frac{5\times 4}{1}=20$

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30 POINTS FOR THIS ONE

satela [25.4K]

Answer:

1. 0.5

2. 3/4

3. -6

4. 5x

Step-by-step explanation:

1. A coefficient is a number that is placed in front of an algebraic term/variable in an expression or equation.

The decimal coefficient is therefore 0.5. It is the coefficient of x.

2. The fractional coefficient is 3/4. It is the coefficient of y.

3. The negative coefficient is -6. It is the coefficient of b.

4. A like term is one which contains the same type of terms as another term.

3x is a like term of 5x, because it contains a numerical term and an algebraic term to the power of 1.

5 0

2 years ago

On average, a student’s grade decreases 2 points for every 30 minutes they spend playing Fortnite. what is the y intercept , slo

aliina [53]

Answer:

y intercept is 0,0 Slope is 15 equation is y=15x

6 0

2 years ago

Which of the following equals 140 to nearest 10

jarptica [38.1K]

Which of the following equals 140 to nearest 10
A.134

B.145

C.136

D.146

6 0

2 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 :

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