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2 years ago

12

What is the value of the expression?

1 answer:

Lady bird [3.3K]2 years ago

5 0

Answer:

690

Step-by-step explanation:

(323+1715)-(1135+213)

(2038)-(1348)

=690

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Can someone help me divide fractions?

GalinKa [24]

Jehdhhdnro wiwjjebeuii ejwjwhhwuabhe

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3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

What is an equation of the line that passes through the point (- 5, - 6) and is parallel to the line 4x - 5y = 35

Elena L [17]

Answer:

4x - 5y = 10

Step-by-step explanation:

Any line parallel to 4x - 5y = 35 will have the same equation EXCEPT that the constant will be different.

Starting with 4x - 5y = 35, replace x with the given x-coordinate -5 and the given y-coordinate -6, and finally the given 35 with the constant C:

4(-5) - 5(-6) = C, or

-20 + 30 = C. Thus, C = 10, and the equation of the new line is

4x - 5y = 10

7 0

3 years ago

PLEASE SHOW WORK

CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

$V = \dfrac{\pi}{3}r^2h$

Taking the derivative of V with respect to time, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)$

$\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$

Since r is always equal to h, we can set

$\dfrac{dr}{dt} = \dfrac{dh}{dt}$

so that our expression for dV/dt becomes

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)$

$\:\:\:\:\:\:\:= \pi r^2\dfrac{dh}{dt}$

Solving for dh/dt, we get

$\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}$

$\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})$

$\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}$

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3 years ago

Can someone help me with this?Can someone help me with this!!!

Elenna [48]

Answer: first fish is -4 and the other is -3

-3 is greater

Step-by-step explanation:

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6 0

2 years ago

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