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2 years ago

11

What is 522,000 rounded to the nearest tenth.

1 answer:

tiny-mole [99]2 years ago

3 0

522,000. This is most likely a question meant to trick you.

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I need help with this problem. It's urgent

Maslowich

Answer:

32 servings

Step-by-step explanation:

There are eight eighths in a cup. Multiply 8 by 4 and you get 32.

4 0

2 years ago

Read 2 more answers

Factorise :<br>ab(x^2 - y^2) + (a^2 - b^2)xy = 0

yawa3891 [41]

Step-by-step explanation:

abx^2-aby^2+xya^2-xyb^2

abx^2+xya^2-aby^2-xyb^2

ax(bx+ya)-yb(ay+xb)

ax(bx+ya)-yb(xb+ay)

(ax-yb) (bx+ya)

7 0

1 year ago

Show that (2xy3 + cos x)dx + (3x2 y2 − sin y)dy = 0 is exact, and find the solution. Find c if y(0) = π

Licemer1 [7]

We're looking for a solution of the form $F(x,y)=C$ . By the chain rule, this solution should have total differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

and the equation is exact if the mixed second-order partial derivatives of $F$ are equal, i.e. $\frac{\partial^2F}{\partial x\partial y}=\dfrac{\partial^2F}{\partial y\partial x}$ .

The given ODE is exact, since

$\dfrac{\partial(2xy^3+\cos x)}{\partial y}=6xy^2$

$\dfrac{\partial(3x^2y^2-\sin y)}{\partial x}=6xy^2$

Then

$\dfrac{\partial F}{\partial x}=2xy^3+\cos x\implies F(x,y)=x^2y^3+\sin x+f(y)$

$\dfrac{\partial F}{\partial y}=3x^2y^2-\sin y=3x^2y^2+\dfrac{\mathrm df}{\mathrm dy}$

$\dfrac{\mathrm df}{\mathrm dy}=-\sin y\implies f(y)=\cos y+C$

$\implies x^2y^3+\sin x+\cos y=C$

With $y(0)=\pi$ , we get

$\cos\pi=C\implies C=-1$

$\implies\boxed{x^2y^3+\sin x+\cos y=-1}$

6 0

2 years ago

True or false: the equation tan^2x+1=sec^2x

zhenek [66]

ANSWER

True

EXPLANATION

The given trigonometric equation is:

${ \tan}^{2} x + 1 = { \sec}^{2} x$

We take the LHS and simplify to arrive at the RHS.

${ \tan}^{2} x + 1 = \frac{{ \sin}^{2} x}{{ \cos}^{2} x} + 1$

Collect LCM on the right hand side to get;

${ \tan}^{2} x + 1 = \frac{{ \sin}^{2} x + {\cos}^{2} x}{{ \cos}^{2} x}$

This implies that

${ \tan}^{2} x + 1 = \frac{1}{{ \cos}^{2} x} .$

${ \tan}^{2} x + 1 = {( \frac{1}{ \cos(x) }) }^{2}$

${ \tan}^{2} x + 1 = { \sec}^{2} x$

This identity has been verified .Therefore the correct answer is true.

3 0

2 years ago

Help me, please !!!!!!!!!

Flura [38]

They did make the same color

7 0

2 years ago