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Yakvenalex [24]

3 years ago

7

Find the distance between (0, -5) & (18, -10).

1 answer:

Otrada [13]3 years ago

8 0

Answer is D
because √(18-0)²-(-10--5)²=√349

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Which situation is repesented by the graph

forsale [732]

Answer:

answer is A i think im not sure

Step-by-step explanation:

i dont know ask your teacher

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3 years ago

Which expression is equivalent to StartFraction 28 p Superscript 9 Baseline q Superscript negative 5 Baseline Over 12 p Superscr

lara31 [8.8K]

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its c

Step-by-step explanation:

7 0

4 years ago

Read 2 more answers

_11] If the radius of a circle is 5, what is the circumference?

ArbitrLikvidat [17]

Answer:

Step-by-step explanation:

A circle with a radius of 5 units has a circumference of 31.416 units.

8 0

3 years ago

Write the equation of the line that passes through (8,1) and is perpendicular to the line whose equation is y=−4x+2. Show all wo

scoundrel [369]

Answer:

$\displaystyle y = \frac{1}{4}x - 1$

Step-by-step explanation:

1 = ¼[8] + b

2

$\displaystyle -1 = b \\ \\ y = \frac{1}{4}x - 1$

* Perpendicular Lines have OPPOSITE MULTIPLICATIVE INVERSE <em>RATE OF CHANGES</em> [<em>SLOPES</em>], so −4 becomes ¼.

I am joyous to assist you anytime.

3 0

4 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

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

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

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

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

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

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

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

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

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

3 years ago