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3 years ago

9

Each year, roughly `10^{6\ }`computer programmers each make about $`10^{5}`. How much money is this all together? Express your a

nswer both as a power of 10 and as a dollar amount.

1 answer:

mina [271]3 years ago

3 0

Answer:

69

Step-by-step explanation:

69e420+60532598423534543

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What is the domain of the function<br> Y=2/x-5

aksik [14]

Answer:

x ≥ 5

Step-by-step explanation:

Find the domain by finding where the equation is defined.

Interval Notation:

(5, ∞ )

Set-Builder Notation:

{x | x ≥ 5}

So, the answer is : x ≥ 5

(PLEASE MARK ME AS BRAINLIEST!!!)

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2 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

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

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

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

(c) Parameterize $C$ by

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

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

Which equation represents the graph below?<br><br> A. y=3x-2<br> B.y=-2x+3<br> C.y=3x-2<br> D.2x+3

KatRina [158]

Answer:

d

Step-by-step explanation:

5 0

3 years ago

A ball has a radius of 8 cm. What is the volume of the ball? Use 3.14 for pie Round your answer to the nearest tenth of a cubic

AveGali [126]

The volume of the ball is 33.5 cubic centimeter.

<u>Step-by-step explanation:</u>

It is given that, a ball has a radius of 8 cm.

We know that, the ball is in shape of the sphere.

The question is asked to find the volume of the ball.

To find the volume of the ball, you need to use the formula for volume of the sphere.

Volume of the sphere = (4/3)πr³

where,

π has the default value of 3.14
r is the radius of the ball.

Volume of the ball = (4/3)× 3.14× 8

⇒ 33.49

⇒ 33.5 (rounded to the nearest tenth)

⇒ 33.5 cubic centimeter.

∴ The volume of the ball is 33.5 cubic centimeter.

6 0

3 years ago

What is the number for PI?

DerKrebs [107]

3.14

This is typically what you would use but there are several other numbers after 3.14

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3 years ago