Ask question

Ask question

All categories

3 years ago

13

Explain why the absolute value of a number is never a negative number ?

2 answers:

omeli [17]3 years ago

4 0

The absolute value of a number is never a negative number because <span>absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.</span>

bija089 [108]3 years ago

3 0

Because when you find the absolute value you're finding the distance a number is from zero without considering its sign ( + or -).

You might be interested in

Find the circumference of a circle<br> with a diameter of 45.

sleet_krkn [62]

Answer:

141.4 centimeters

Step-by-step explanation:

circumference = diameter * pi c = d * pi c = 45 * 3.1416 c = 141.4 (rounded)

7 0

2 years ago

Read 2 more answers

Ariel is working at a meat-packing plant 5 nights a week. Her regular wage is $11 an hour. She earns time and a half for any ove

svetlana [45]

Answer:

$11 would be earned

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

What is the slope of the line with equation 4x + 2y = 8?

Papessa [141]

Y=2x+4
Hope this helps :D

8 0

3 years ago

Read 2 more answers

Plz help urgent question below

zhannawk [14.2K]

A is the answer. Hope this helps

6 0

3 years ago

Find the general solution of the equation: (y^2-2xy)dx + (2xy-x^2)dy = 0

lys-0071 [83]

The ODE is exact, since

$\dfrac{\partial(y^2-2xy)}{\partial y}=2y-2x$

$\dfrac{\partial(2xy-x^2)}{\partial x}=2y-2x$

so there is solution $f(x,y)=C$ such that

$\dfrac{\partial f}{\partial x}=y^2-2xy$

$\dfrac{\partial f}{\partial y}=2xy-x^2$

Integrating both sides of the first PDE wrt $x$ gives

$f(x,y)=xy^2-x^2y+g(y)$

Differentiating both sides wrt $y$ gives

$\dfrac{\partial f}{\partial y}=2xy-x^2+\dfrac{\mathrm dg}{\mathrm dy}=2xy-x^2$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=0\implies g(y)=C$

Then the solution to the ODE is

$f(x,y)=\boxed{xy^2-x^2y=C}$

# # #

Alternatively, we can see that the ODE is homogeneous, since replacing $x\to tx$ and $y\to ty$ reduces to the same ODE:

$((ty)^2-2(tx)(ty))\,\mathrm d(tx)+(2(tx)(ty)-(tx)^2)\,\mathrm d(ty)=0$

$t^3(y^2-2xy)\,\mathrm dx+t^3(2xy-x^2)\,\mathrm dy=0$

$(y^2-2xy)\,\mathrm dx+(2xy-x^2)\,\mathrm dy=0$

This tells us we can solve by substituting $y(x)=xv(x)$ , so that $\mathrm dy(x)=x\,\mathrm dv(x)+v(x)\,\mathrm dx$ , and the ODE becomes

$(x^2v^2-2x^2v)\,\mathrm dx+(2x^2v-x^2)(x\,\mathrm dv+v\,\mathrm dx)=0$

$v(v-2)\,\mathrm dx+(2v-1)(x\,\mathrm dv+v\,\mathrm dx)=0$

$3v(v-1)\,\mathrm dx+x(2v-1)\,\mathrm dv=0$

which is separable as

$\dfrac{1-2v}{3v(v-1)}\,\mathrm dv=\dfrac{\mathrm dx}x$

Integrating both sides gives

$-\dfrac13\ln|v(1-v)|=\ln|x|+C$

$v(1-v)=\dfrac C{x^3}$

and solving in terms of $y(x)$ ,

$\dfrac yx\left(1-\dfrac yx\right)=\dfrac C{x^3}$

$xy(x-y)=C$

$\boxed{x^2y-xy^2=C}$

6 0

2 years ago