All categories

3 years ago

Dose anyone know the answer to this question?

Mathematics

2 answers:

natima [27]3 years ago

8 0

Answer:

$\boxed{\sf -3}$

Step-by-step explanation:

The denominators are same. Add fractions.

$\displaystyle \sf \frac{-3k+6}{k-2}$

Factor out the expression in the numerator.

$\displaystyle \sf \frac{-3(k-2)}{k-2}$

Cancel common terms.

$\displaystyle \sf \frac{-3}{1}=-3$

RideAnS [48]3 years ago

4 0

Answer:

-3

Step-by-step explanation:

-3k / ( k-2) + 6 / ( k-2)

since the denominators are the same we can add the numerators

(-3k+6) / ( k-2)

Factor out -3

-3 ( k-2)/ ( k-2)

Cancel like terms

-3

You might be interested in

What subtraction sentence does this show?

harina [27]

Ummmmmmmmmmmmm
Mm A is correct I think. The question was unclear but A is the only correct function

5 0

2 years ago

Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exact<br><br> C=.

Tpy6a [65]

Answer with Step-by-step explanation:

The given differential equation is

$(2x+5y)dx+(5x-4y)dy=0$

Now the above differential equation can be re-written as

$P(x,y)dx+Q(x,y)dy=0$

Checking for exactness we should have

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$

$\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5$

$\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5$

As we see that the 2 values are equal thus we conclude that the given differential equation is exact

The solution of exact differential equation is given by

$u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)$

The value of $\phi (y)$ can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)

$\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c$