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2 years ago

5

Winston bought an encyclopedia priced at $42. Shipping and handling cost anadditional 30% of the price. What was the total cost

of the encyclopedia, in cluding shipping and handling?

1 answer:

Sergio039 [100]2 years ago

3 0

Answer:

42.3

Sorry I haven't been answering your questions I'm just behind on homework.

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2x+3(x-2)-3(x-6) figuring for X

emmainna [20.7K]

Answer:

2x+12

Step-by-step explanation:

2x+3(x-2)-3(x-6)

2x+3x-6-3x+18

2x-6+18

2x+12

6 0

2 years ago

Read 2 more answers

Adam was going to sell all of his stamp collection to buy a video game. After selling half of them he changed his mind and did n

Roman55 [17]

He would have 21 before he bought the 10 stamp because 31-10=21

6 0

3 years ago

Which of the sets of equations represent the two lines graphed?

Oksi-84 [34.3K]

Answer:

B) y = x + 2 and y = -x - 4

Step-by-step explanation:

Let the equation of a straight line with x-intercept 'a' and y-intercept 'b' be

$\frac{x}{a} + \frac{y}{b} = 1$

The line with positive slope has x-intercept a=-2 and y-intercept b=2.

Its equation is:

$\frac{x}{ - 2} + \frac{y}{2} = 1$

Multiply through by 2

$- x + y = 2$

Solve for y,

$\boxed {y = x + 2}$

For the line with a negative slope,

the x-intercept is a=-4 and the y-intercept is b=-4

Its equation is

$\frac{x}{ - 4} + \frac{y}{ - 4} = 1$

Multiply through by -4

$x + y = - 4$

Solve for y

$\boxed {y = - x - 4}$

4 0

2 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

3 0

3 years ago

The ordered pair (a, b) gives the location of point P on the coordinate plane. The values of a and b have

Mama L [17]

Answer:

Since a and b = (a, b) = (x, y) have the same sign.....P will lie either in the 1st quadrant of the Cartesian plane or the 3rd quadrant of the Cartesian plane.

6 0

2 years ago