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3 years ago

IF YOU ANSWER CORRECTLY I WILL MAKE ANOTHER QUESTION ADN GIVE YOU 20 POINTS

Mathematics

2 answers:

Montano1993 [528]3 years ago

6 0

Answer:

Im not sure what the question is but if its what the relation is between the boxes the first one would be y = x+5 the second one would be y=x-4 the third would be y= x+3 and the fourth would be y= x times 2

Step-by-step explanation:

Korolek [52]3 years ago

4 0

Step-by-step explanation:

Pretty sure it is Linear Function, just graphed it.

Linear Function: y=mx+b when m is slope and b is y-intercept.

From the graph that I drew, the first one. y-intercept is at 5 and the slope is 1.

So the first one is y=x+5.

For the second one, x-intercept is at 4. That means y-intercept is at -4.

So the second one is y=x-4

If you draw both lines all at once, you will notice that both lines don't have/share interception. That means there are no answers for the equation.

y=x+5 — 1

y=x-4 — 2

Substitute 1 in 2

x+5=x-4

x-x+5+4=0

9=0

The equation doesn't have answers.

And I cant see the third one anyways, because I am on phone.

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This years “Dream of a lifetime” lottery sold 50,000 tickets and gave away 5,000 prizes including one grant prize

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Answer:

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Step-by-step explanation:

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3 years ago

Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8

Ivanshal [37]

It looks like the differential equation is

$\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}$

Factorize the right side by grouping.

$xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)$

$xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)$

Now we can separate variables as

$\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx$

Integrate both sides.

$\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx$

$\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx$

$\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}$

You could go on to solve for $y$ explicitly as a function of $x$ , but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.