All categories

2 years ago

Which is the equation for ---

Mathematics

1 answer:

ratelena [41]2 years ago

7 0

Equation of the line:
y=mx+b
Ordinate at the origin (where the line cut the y-axis): b=1

B=(-5,5)=(xb,yb)→xb=-5, yb=5
C=(0,1)=(xc,yc)→xc=0, yc=1
m=(yc-yb)/(xc-xb)
m=(1-5)/(0-(-5))
m=(-4)/(0+5)
m=(-4)/(5)
m=-(4/5)

y=mx+b
y=-(4/5)x+1
Multiplying the equation by 5:
5[y=-(4/5)x+1]
5y=-4x+5
Adding 4x both sides of the equation:
5y+4x=-4x+5+4x
4x+5y=5

Answer: The equation for BC is: Third option 4x+5y=5

You might be interested in

(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

Please help!!!!!!! it would be very appreciated

Pani-rosa [81]

Answer:

x = 17.5 and y = 35

Step-by-step explanation:

2x + 10 = y + 10 ( alternate angles )

subtract 10 from both sides

2x = y ⇒ 6x = 3y → (1)

the sum of the 3 angles in the triangle = 180°

6x + y + 10 + 30 = 180 ( replace 6x by 3y )

3y + y + 40 = 180

4y + 40 = 180 ( subtract 40 from both sides )

4y = 140 ( divide both sides by 4 )

y = 35

substitute y = 35 into (1)

2x = 35 ( divide both sides by 2 )

x = 17.5

5 0

2 years ago

Simplify √14q times 2√4q

DerKrebs [107]

4q√14 hope this help

5 0

2 years ago

Read 2 more answers

Parallelogram WXYZ is shown on the coordinate plane.

gogolik [260]

The reflection transformation of the given diagram means that; W' X' is parallel to Z' Y'

<h3>How to Interpret Transformation Reflection?</h3>

Reflection over y-axis is a reflection or flip over the y-axis where the y-axis is the line of reflection used. The formula for this is: (x,y) →(−x,y) ( x , y ) → ( − x , y ) .

To reflect an equation over the y-axis, simply multiply the input variable by -1: y=f(x)→y=f(−x) y = f ( x ) → y = f ( − x )

Thus, looking at the given image and applying the transformation rule above, we can say that the reflection of parallelogram WXYZ across the y-axis would mean that W' X' is parallel to Z' Y'

Read more about Transformation Reflection at; brainly.com/question/1908648

#SPJ1

5 0

1 year ago

Yvonne was asked to construct a line through P that is perpendicular to the given line segment AB for homework.

Oxana [17]

Answer: C

Step-by-step explanation:

Place the compass on point and mark an arc above AB that goes through P, and a similar arc below AB.

I just did the CYU and this was correct

6 0

2 years ago