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3 years ago

Slope intercept Of equation

Mathematics

2 answers:

dimulka [17.4K]3 years ago

8 0

Y=Mx+b
B is the y inter and m is the x inter

dedylja [7]3 years ago

4 0

Answer:

In general, the slope intercept form assumes the formula: y = mx + b. m is the slope (lesson on slope ) mnemonic : 'm' means 'move' b is the y -intercept ( lesson on the y-intercept) mnemonic : 'b' means where the line begins.

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Solve P= 2B + 2M for B

kykrilka [37]

Answer:

B = P/ 2 − M

Step-by-step explanation:

8 0

2 years ago

You buy 1.75 pounds of apples and 2.05 pounds of oranges. Apples and oranges sell for the same price per pound. The total cost,

dlinn [17]

Answer:

$1.75c+2.05c=4.18\\\\Per\ pound\ cost=c=\$1.1$

Step-by-step explanation:

$Cost\ of\ per\ pound\ apple=Cost\ of\ per\ pound\ oranger=c\\\\$

Cost of Apples:

$Amount\ of\ apples=1.75\ pounds\\\\Cost\ of\ apples=Amount\times per\ pound\ cost\\\\Cost\ of\ apples=1.75\times c\\\\Cost\ of\ apples=1.75c$

Cost of oranges:

$Amount\ of\ oranges=2.05\ pounds\\\\Cost\ of\ oranges=Amount\times per\ pound\ cost\\\\Cost\ of\ oranges=2.05\times c\\\\Cost\ of\ oranges=2.05c$

Total cost:

$Total\ cost\ of\ fruits=cost\ of\ apple+cost\ of\ oranges\\\\Total\ cost\ of\ fruits=1.75c+2.05c\\\\After\ coupon\ cost=\$ 3.43\\\\ Original\ cost=After\ coupon\ cost+coupon\ amount\\\\ Original\ cost=3.43+0.75=4.18$

$1.75c+2.05c=4.18\\\\3.8c=4.18\\\\c=\frac{4.18}{3.8}\\\\c=1.1\\\\Per\ pound\ cost\ of\ fruits=\$1.1$

4 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 :

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