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2 years ago

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1. What is the general form of the equation for this line y-5= -3(x + 4)?

2 answers:

ivanzaharov [21]2 years ago

8 0

Answer:

I think its B

Serggg [28]2 years ago

4 0

It’s is c 3x + y= -7

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Which statement is true?

Dovator [93]

All squares are rectangles

6 0

3 years ago

Read 2 more answers

Use the Laplace transform and inverse Laplace transform to solve in the initial value problem y''-3y'+2y=4t-6 y(0)=1, y'(0)=2

Vesnalui [34]

Denote by $Y(s)$ the Laplace transform of $y(t)$ .

$y''-3y'+2y=4t-6$

Take the Laplace transform of both sides:

$(s^2Y(s)-sy(0)-y'(0))-3(sY(s)-y(0))+2Y(s)=\dfrac4{s^2}-\dfrac6s$

Solve for $Y(s)$ :

$(s^2Y(s)-s-2)-3(sY(s)-1)+2Y(s)=\dfrac4{s^2}-\dfrac6s$

$(s^2-3s+2)Y(s)=\dfrac4{s^2}-\dfrac6s+s-1$

$Y(s)=\dfrac{4-6s-s^2+s^3}{s^2(s^2-3s+2)}$

$Y(s)=\dfrac{4-6s-s^2+s^3}{s^2(s-1)(s-2)}$

Decompose the right side into partial fractions: we're looking for constants $a_1,\ldots,a_4$ such that

$\dfrac{4-6s-s^2+s^3}{s^2(s-1)(s-2)}=\dfrac{a_1}s+\dfrac{a_2}{s^2}+\dfrac{a_3}{s-1}+\dfrac{a_4}{s-2}$

$4-6s-s^2+s^3=a_1s(s-1)(s-2)+a_2(s-1)(s-2)+a_3s^2(s-2)+a_4s^2(s-1)$

Expanding on the right side gives

$2a_2+(2a_1-3a_2)s-(3a_1-a_2+2a_3+a_4)s^2+(a_1+a_3+a_4)s^3$

and matching up coefficients gives the system

$\begin{cases}2a_2=4\\2a_1-3a_2=-6\\3a_1-a_2+2a_3+a_4=1\\a_1+a_3+a_4=1\end{cases}\implies a_1=0,a_2=2,a_3=2,a_4=-1$

So we have

$Y(s)=\dfrac2{s^2}+\dfrac2{s-1}-\dfrac1{s-2}$

and taking the inverse transform of both sides is trivial, giving

$\boxed{y(t)=2t+2e^t-e^{2t}}$

4 0

3 years ago

Let f(x)=2x-1 and g(x)=4-x^2. Find (f o g)(x).

stepan [7]

$fog(x)=f(g(x)) \\ \\ \\ =2.(4-x^2)-1 \\ \\ \\ = -2x^2+7 \\ \\ \\ GOOD \quad LUCK$

6 0

3 years ago

HELP !!!!! ASAP 4 ENDGENUITY Let the graph of g be a translation 5 units to the right followed by a translation 7 units down of

Kruka [31]

A translation of $h$ units upwards and $q$ units to the right is represented by the transformation

$(x,y)\rightarrow(x-q,y-h)$

So, to translate 5 right and 7 down, we need

$(x,y)\rightarrow(x-5,y+7)$

Now we just need to plug x=x-5 and y=y+7 in our equation (remember f(x)=y):

$y+7=|x-5|\\y=|x-5| - 7$

The answer is

$\boxed{g(x) = |x-5|-7}$

6 0

2 years ago

Let f(x) = x-9<br> Find (f o f)(x) and its domain<br> (f o f)(x) =

Oksi-84 [34.3K]

Answer:

Step-by-step explanation:

6 0

3 years ago