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maxonik [38]
3 years ago
5

Which values for A and B will create infinitely many solutions for this system of equations

Mathematics
1 answer:
Ratling [72]3 years ago
7 0

Answer:

A = - 2 and B = - 8

Step-by-step explanation:

A = - 2 and B = - 8 will create infinitely many solutions for this system of equations

When you add both equations together and both sides are equal then the systems have infinitely many solutions

-2x - y = 8

2x + y = -8

--------------------Add

0 = 0

Both sides are equal 0

Infinitely many solutions

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A ship is moving at a speed of 30 km/h parallel to a straightshoreline. The ship is 6 km from the shore and it passes alighthous
natita [175]

Answer:

A)  S = f(d) = √(36 + d²)

B)  d = g(t) = 30t

C)  f o g = √(36 + 900t²)

Explanation:  

A) Express the distance s between the lighthouse and the ship as a function of d, the distance the ship has traveled since noon;that is, find f so that s = f(d)

Refer to the attached image and observe that ship starts moving from point I towards point N which represents the position of ship at noon.

From point I to point N ship covers the distance d.

From point I to point L (lighthouse) the distance is S.

A triangle is formed and we want the distance S so we will apply the Pythagorean theorem

IL² = NL² + IN²

S² = 6² + d²

S = √(36 + d²)

Therefore, the function is S = f(d) = √(36 + d²)

B) Express d as a function of t, the time elapsed since noon;that is, find g, so that d = g(t)

Since the ship is moving at a speed of 30 km/h and let t represents the time taken to move from I to N then

d = 30t

or

d = g(t) = 30t

C) Find f o g. What does this function represent?

Since we have already found f(d) and g(t) we can find f o g.

f(d) = √(36 + d²) and d = g(t) = 30t

substitute d = g(t) = 30t into f(d)

f o g = √(36 + (g(t))²)

f o g = √(36 + (30t)²)

f o g = √(36 + 900t²)

This function simply represents the distance from point I to point N to point L which is the longer distance from the initial position of ship to the lighthouse.

7 0
4 years ago
Guys plzzz helpp me this gotta be done now!!
Bess [88]

Answer:

I think it's C.

Step-by-step explanation:

plz mark brainliest

5 0
3 years ago
Amelia ,Hayden and Sophie did a test the total for the test was 75 marks Amelia got 56% of 75 marks Hayden got 8/15 0f the 75 ma
boyakko [2]

Step-by-step explanation:

Amelia got 56% of 75 so

(56/100)*75=(56*75)/100=4200/100=42

Hayden got 8/15 of 75 so

(8/15)*75=(8*75)/15=600/15=40

and Sophie that got 43 out of 75

so we see that Sophie got the highest points.

6 0
3 years ago
Read 2 more answers
(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 <em>µ(x, y)</em> 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 <em>µ(x, y)</em> = <em>µ(x)</em> be independent of <em>y</em>, then <em>∂µ/∂y</em> = 0 and we can solve for <em>µ</em> :

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 <em>F(x, y)</em> = <em>C</em>. 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 <em>x</em> :

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

Differentiate both sides of this with respect to <em>y</em> :

\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
Sofia raised $140 for charity. Her mom gave her some money to get started, and her friends donated the rest of the money Dominic
Katen [24]

Answer:

Step-by-step explanation:

you add 75 plus 30 which is 105 and subtract that by 140 which is 35

3 0
3 years ago
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