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3 years ago

Given: f(x) = xand g(x) = x +1, find g[f(-2)]. 1 5 -3

Mathematics

2 answers:

AleksAgata [21]3 years ago

7 0

After all the breaking down I would say it’s negative one (-1)

notsponge [240]3 years ago

3 0

Answer:

-1

Step-by-step explanation:

f(x) = x

g(x) = x +1,

g[f(-2)].

First find f(-2) = -2

Then substitute into g(x)

g(-2) = -2+1

=-1

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Find the angle of arc BC

amid [387]

Answer:

96°

Step-by-step explanation:

The measure of arc EC is twice the measure of ∠DEC, so is ...

arc EC = 2×68° = 136°

The measure of arc BC is found from the sum of arcs around a circle.

BC +CE +EB = 360° = BC +136° +128°

BC = 96° . . . . . subtract 264°

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3 years ago

Solve for the value of x and y when 3x+2y=24 , x+y=9<br> step wise explanation plz

OLga [1]

Answer:

{x,y} = {2,9}

Step-by-step explanation:

Solve equation [2] for the variable y

[2] 2y = -6x + 30

[2] y = -3x + 15

// Plug this in for variable y in equation [1]

[1] 3x + 2•(-3x+15) = 24

[1] -3x = -6

// Solve equation [1] for the variable x

[1] 3x = 6

[1] x = 2

// By now we know this much :

x = 2

y = -3x+15

// Use the x value to solve for y

y = -3(2)+15 = 9

6 0

2 years ago

Read 2 more answers

Six friends shared 79 trading cards and kept a record of how they did it. What does 13 represent? the answers are- The number of

castortr0y [4]

Answer: The number of trading cards each friend ended up with

Step-by-step explanation:

Given that:

Total Number of trading cards = 79

Number of friends = 6

Number of trading cards each friend got of shared equally :

Total Number of trading cards / number of friends

= 79 / 6

= 13.166

Highest number of cards got by each = 13

Number left : 79 - (13 * 6) = 1

Hence, 13 represents the number of trading cards each friend ended up with

7 0

3 years ago

Solve the system of equations using substitution. Show your work. <br> y= 2x - 10<br> y= 4x - 8

Semenov [28]

Y=2x-10
Y=4x-8
U substitute it and it turns into 2x-10 =4x-8
U get x=-1
Y=4*(-1)-8
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If u have a phone u can use photo math or if u don’t there’s a thing called mathpapa

4 0

3 years ago

Calculus 3 help please.

Reptile [31]

I assume each path $C$ is oriented positively/counterclockwise.

(a) Parameterize $C$ by

$\begin{cases} x(t) = 4\cos(t) \\ y(t) = 4\sin(t)\end{cases} \implies \begin{cases} x'(t) = -4\sin(t) \\ y'(t) = 4\cos(t) \end{cases}$

with $-\frac\pi2\le t\le\frac\pi2$ . Then the line element is

$ds = \sqrt{x'(t)^2 + y'(t)^2} \, dt = \sqrt{16(\sin^2(t)+\cos^2(t))} \, dt = 4\,dt$

and the integral reduces to

$\displaystyle \int_C xy^4 \, ds = \int_{-\pi/2}^{\pi/2} (4\cos(t)) (4\sin(t))^4 (4\,dt) = 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt$

The integrand is symmetric about $t=0$ , so

$\displaystyle 4^6 \int_{-\pi/2}^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \,dt$

Substitute $u=\sin(t)$ and $du=\cos(t)\,dt$ . Then we get

$\displaystyle 2^{13} \int_0^{\pi/2} \cos(t) \sin^4(t) \, dt = 2^{13} \int_0^1 u^4 \, du = \frac{2^{13}}5 (1^5 - 0^5) = \boxed{\frac{8192}5}$

(b) Parameterize $C$ by

$\begin{cases} x(t) = 2(1-t) + 5t = 3t - 2 \\ y(t) = 0(1-t) + 4t = 4t \end{cases} \implies \begin{cases} x'(t) = 3 \\ y'(t) = 4 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{3^2+4^2} \, dt = 5\,dt$

and

$\displaystyle \int_C x e^y \, ds = \int_0^1 (3t-2) e^{4t} (5\,dt) = 5 \int_0^1 (3t - 2) e^{4t} \, dt$

Integrate by parts with

$u = 3t-2 \implies du = 3\,dt \\\\ dv = e^{4t} \, dt \implies v = \frac14 e^{4t}$

$\displaystyle \int u\,dv = uv - \int v\,du$

$\implies \displaystyle 5 \int_0^1 (3t-2) e^{4t} \,dt = \frac54 (3t-2) e^{4t} \bigg|_{t=0}^{t=1} - \frac{15}4 \int_0^1 e^{4t} \,dt \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} e^{4t} \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \frac54 (e^4 + 2) - \frac{15}{16} (e^4 - 1) = \boxed{\frac{5e^4 + 55}{16}}$

$\begin{cases} x(t) = 3(1-t)+t = -2t+3 \\ y(t) = (1-t)+2t = t+1 \\ z(t) = 2(1-t)+5t = 3t+2 \end{cases} \implies \begin{cases} x'(t) = -2 \\ y'(t) = 1 \\ z'(t) = 3 \end{cases}$

with $0\le t\le1$ . Then

$ds = \sqrt{(-2)^2 + 1^2 + 3^2} \, dt = \sqrt{14} \, dt$

and

$\displaystyle \int_C y^2 z \, ds = \int_0^1 (t+1)^2 (3t+2) \left(\sqrt{14}\,ds\right) \\\\ ~~~~~~~~ = \sqrt{14} \int_0^1 \left(3t^3 + 8t^2 + 7t + 2\right) \, dt \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 t^4 + \frac83 t^3 + \frac72 t^2 + 2t\right) \bigg|_{t=0}^{t=1} \\\\ ~~~~~~~~ = \sqrt{14} \left(\frac34 + \frac83 + \frac72 + 2\right) = \boxed{\frac{107\sqrt{14}}{12}}$

8 0

1 year ago