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3 years ago

13

Simplify. 5^-1(3^-2)

2 answers:

Rina8888 [55]3 years ago

6 0

The answer is 0.002 Hope this helps

wariber [46]3 years ago

4 0

The answer is zero hope that helps

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Which point is a solution to the equation 2x - y = 4?

Natasha_Volkova [10]

Answer:

2<em>x= 1 /2 y+ 2</em>

Step-by-step explanation:

Solve for x.

<em>2x−y=4</em>

Add y to both sides.

<em>2x−y+y=4+y </em>

<em>2x=y+4</em>

Divide both sides by 2.

<em>2x /2 = y+4 /2</em>

<em>x= 1/ 2 y+2</em>

5 0

3 years ago

Which two lines are parallel?

ddd [48]

For lines to be parallel they must have the same slope. If:

y=mx+b

For any line to be parallel to the above it must also have the same value for m in that line.

4 0

3 years ago

3x+4y=12 6x+2y=6 rewrite each equation and soap intercept form

miskamm [114]

Move 4y to the front so it should be 4y=12-3x
Divide the equation like how I did in the picture above then you should have y=3-3/4x
Then you switch it to y=-3/4x+3 and that's your answer.

Now for 6x+2y=6

Move 2y to the front so it looks like 2y=6-6x
Then you divide the equation by 2 like in the photo above and that should give you y=3-3x
Finally you switch the equation to y=-3x+3 because we are using the Slope Intercept Form. (Y=mx+b). That's your answer.

Sorry if my handwriting is bad.

5 0

3 years ago

Which point represents the square root of 182

pentagon [3]

the square root of 182 is 13.4907375632

hope it helps you

mark barinest

3 0

3 years ago

Use the variation of parameters method to solve the DR y" + y' - 2y = 1

postnew [5]

Answer:

$y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

$y"+y'-2y=1$

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

$D^2+D-2\ =\ 0$

$=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}$

$=>\ D=\ -2\ or\ 1$

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

$y_c(t)\ =\ C_1e^{-2t}+C_2e^t$

Let's assume that

$y_1(t)=e^{-2t}$ $y_2(t)=e^t$

$=>\ y'_1(t)=-2e^{-2t}$ $y'_2(t)=e^t$

and g(t)=1

Now, the Wronskian can be given by

$W=y_1(t).y'_2(t)-y'_1(t).y_2(t)$

$=e^{-2t}.e^t-e^t(-e^{-2t})$

$=e^{-t}+2e^{-t}$

$=3e^{-t}$

Now, the particular solution can be given by

$y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}$

$=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}$

$=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}$

$=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}$

$=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

Now, the complete solution of the given differential equation can be given by

$y(t)\ =\ y_c(t)+y_p(t)$

$=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}$

5 0

3 years ago