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3 years ago

Evaluate: 4vw/-8 -|v| when v=-2 and w=3

Mathematics

2 answers:

Kaylis [27]3 years ago

6 0

Answer: 1

Step-by-step explanation:

{[4x(-2)x(3)]/(-8)}-2=1

Marta_Voda [28]3 years ago

3 0

Answer:

Step-by-step explanation:

4*-2*3=-24 -24/-8

reduced=3

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4(2x - 7) - 2x = -10

jarptica [38.1K]

So the right answer is 3.

look at the attached picture

Hope it will help you..

Good luck on your assignment

~pragya

5 0

3 years ago

Read 2 more answers

Question number 2 answer asap please

slava [35]

Answer:

The ordered pair that is the solution to the system lies in Quadrant IV

The x-coordinate of the solution is 9

The y-coordinate of the solution is -8

Step-by-step explanation:

8 0

3 years ago

A stack of playing cards contains 4 jacks, 5 queens, 3 kings, and 3 aces. two cards will be randomly selected from the stack. wh

Kipish [7]

Answer:

1/9.

Step-by-step explanation:

There is a total of 15 cards in the stack.

Prob( Queen is chosen) = 5/15 = 1/3.

The probability of a second queen being chosen is also 1/3

Required probability = 1/3 * 1/3 = 1/9.

7 0

3 years ago

The absolute value function, f(x) = –|x| – 3, is shown. What is the range of the function? all real numbers all real numbers les

Harman [31]

Answer:

all real numbers less than or equal to –3

Step-by-step explanation:

Look at the y-values that the graph shows. There are none greater than -3. Any value of -3 or less is possible.

5 0

3 years ago

Read 2 more answers

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago