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3 years ago

NEED HELPS ASAP WILL GIVE BRAINLIST 100 PTS

Mathematics

2 answers:

Slav-nsk [51]3 years ago

5 0

A term is a single number or variable, in this case, the number 2.

The sum in this equation would be 4x + 2.

The product is the result of a multiplication, in this equation, the result of 5(4x + 2).

The factors are the digits that are being multiplied, in this case 5 & 4x + 2

The quotient is the result of a division, in this case, the result of (4x + 2) / 7y.

The coefficient is a number that is being multiplied by a variable, in this case, 4 & y.

Hope it helped,

BioTeacher101

hodyreva [135]3 years ago

4 0

Hey there!

$\ \ \ \ \ \ \ \ \ \Downarrow \ \Downarrow \Downarrow\ \ \ \ \Downarrow \ \ \ \ \Downarrow

\\ \\ \left[\begin{array}{ccc} \left[\begin{array}{ccc} \left[\begin{array}{ccc}5(4x + 2)/7y\end{array}\right] \end{array}\right] \end{array}\right]$

$(5)=(term) \\ \\ (sum)=(+) \\ \\ (product)= (x) \ or \ (*) \\ \\ (4)=(coefficient) \\ \\ (quotient) = (divided \ by \ symbol) \\ \\ (factor)=(2) \\ \\ \boxed{\boxed{Hope \ this \ helps}}$

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3 years ago

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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

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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

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

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

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

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3 years ago

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Answer:

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