Which three R's are associated with decreasing the world's waste?

For f(x) = 3x - 1 and g(x) = x+2, find (f – g)(x).

kobusy [5.1K]

Well, I bet you want your answer right away! So here it is.

Given f (x) = 3x + 2 and g(x) = 4 – 5x, find (f + g)(x), (f – g)(x), (f × g)(x), and (f / g)(x).

To find the answers, all I have to do is apply the operations (plus, minus, times, and divide) that they tell me to, in the order that they tell me to.

(f + g)(x) = f (x) + g(x)

= [3x + 2] + [4 – 5x]

= 3x + 2 + 4 – 5x

= 3x – 5x + 2 + 4

= –2x + 6

(f – g)(x) = f (x) – g(x)

= [3x + 2] – [4 – 5x]

= 3x + 2 – 4 + 5x

= 3x + 5x + 2 – 4

= 8x – 2

(f × g)(x) = [f (x)][g(x)]

= (3x + 2)(4 – 5x)

= 12x + 8 – 15x2 – 10x

= –15x2 + 2x + 8

\left(\small{\dfrac{f}{g}}\right)(x) = \small{\dfrac{f(x)}{g(x)}}(gf)(x)=g(x)f(x)= \small{\dfrac{3x+2}{4-5x}}=4−5x3x+2

My answer is the neat listing of each of my results, clearly labelled as to which is which.

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f × g ) (x) = –15x2 + 2x + 8

\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}(gf)(x)=4−5x3x+2

Hope I helped! :) If I did not help that's okay.

-Duolingo

7 0

3 years ago

Value of 2x when x = 5

lora16 [44]

Answer: 10

Step-by-step explanation:

My answer is correct because, if x=5 then were just multiplying 2 times 5 since 5 is not on the top like $2^{5}$ because if it was that means we have to multiply 2 by itself 5 times. But since its not a exponent you just multiply 2 times 5. And 2 x 5= 10.

Hope This Helps!

8 0

3 years ago

Read 2 more answers

Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8

Kitty [74]

Substitute $z=\ln x$ , so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)$

Then the ODE becomes

$x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0$
$\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0$

which has the characteristic equation $r^2+1=0$ with roots at $r=\pm i$ . This means the characteristic solution for $y(z)$ is

$y_C(z)=C_1\cos z+C_2\sin z$

and in terms of $y(x)$ , this is

$y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$

From the given initial conditions, we find

$y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1$
$y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8$

so the particular solution to the IVP is

$y(x)=\cos(\ln x)+8\sin(\ln x)$

4 0

3 years ago

Richard ran the route A-B-C-D-E-F-G-H-A. Which of the following paths taken by Richard appear to be perpsndicular to path BC?

Artist 52 [7]

<h3>Answer: B) segment CD</h3>

Perpendicular segments form a 90 degree angle (aka right angle). Segment BC appears to be perpendicular to both AB and CD. I use the phrasing "appears to be" since it's not entirely clear if they actually are perpendicular. The use of a square angle marker would indicate the segments form a 90 degree angle. Such a symbol is missing.

8 0

4 years ago

The museum admission is $14 and a student is 10$ how much will it cost for one class

Dmitriy789 [7]

Well how many students are there in the class??

4 0

3 years ago