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Murljashka [212]
3 years ago
8

Simplify. Assume that no denominator is equal to zero. 4^12/4^5 A. 4^7 B. 4^-7 C. 4^5 D. 1^7

Mathematics
2 answers:
SVETLANKA909090 [29]3 years ago
8 0

Answer:

A

Step-by-step explanation:

\frac{a^{m}}{a^{n}}=a^{m-n}\\\\\frac{4^{12}}{4^{5}}=4^{12-5}\\\\       = 4^{7}

Burka [1]3 years ago
8 0

Answer:

A. 4^7

Step-by-step explanation:

4^12/4^5= 4^(12-5)=4^7

Option A

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How do I factor this? <br> 5x^2y+45x^3y^2
ruslelena [56]

Answer:

5 x^2 y (9 x y + 1)

Step-by-step explanation:

Factor the following:

45 x^3 y^2 + 5 x^2 y

Factor common terms out of 45 x^3 y^2 + 5 x^2 y.

Factor 5 x^2 y out of 45 x^3 y^2 + 5 x^2 y:

Answer: 5 x^2 y (9 x y + 1)

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2 years ago
) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra
artcher [175]

Answer:

y(t)=2e^{3t}(2-5t)

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2  

Using the theorem of the Laplace transform for derivatives, we know that:

\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)

Replacing the initial values y(0)=4, y′(0)=2 we obtain

\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4

and our differential equation (*) gets transformed in the algebraic equation

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0

Solving for Y(s) we get

\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}

Now, we brake down the rational expression of Y(s) into partial fractions

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here  

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}

and

\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}

we know that

\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}

and for the first translation property of the inverse Laplace transform

\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}

and the solution of our differential equation is

\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}

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3 years ago
Help plsssssssss especially on Thursday number 3 pls brainliest
sergiy2304 [10]

Answer:

2. Thursday

A

3. Thursday

mean = 26.4

median = 27

mode = 18, 21, 27, 32, 24, 30, 33

range = 15

4. Thursday

is in the picture

5. Wednesday

2, 1, -12, -23, -24

5, 1, -43, -50, -54

hope dis helps ^-^

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I need help on these two questions
Marina86 [1]

Answer:

1. $60 2. $233.20

Step-by-step explanation:

That is the answer

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2 years ago
Will give brainliest
tester [92]

Answer:

its all sides are equal and diagonal make 90° angle.

so, the answer option c.

4 0
3 years ago
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