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3 years ago

Simplify x3/x A.x2 B. 1/x2

Mathematics

2 answers:

masha68 [24]3 years ago

5 0

X3 is the same as 3x or 3 times x so

3x/x=3

Hunter-Best [27]3 years ago

4 0

Simplify x3/x A.x2 B. 1/x2 - 3705076. ... + 4 f(x) = x2 – 5x + 4 f(x) = x2 + 3x – 4 f(x) = –2x2 + 10x – 8 f(x) = –4x2 – 16x – 1.

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Kyla works as a receptionist. Last week she made $236.40 for working 24 hours. About how many hours would she need to work to ea

kow [346]

Answer:

32.99

Step-by-step explanation:(you might have to round) if you have to round it or.. but this is it because

236.40 divided by 24 is 9.85 so

325 divided by 9.85 is 32.99 (you only need 2 decimal places for these kinds of questions because of repeating decimals).

Hope this helps :)

8 0

3 years ago

Could I get the difference of the meaning?

____ [38]

Absolute value of -3 is three, and then it just regular -3 equals -3 so basically whenever you see at -3 in those brackets or any negative number it means absolute value which will always be it’s positive version That negative number

4 0

3 years ago

A driver intends to complete 2 laps on a track that is exactly 1 mile around at an average speed of 60 mph. He gets off to a slo

andrey2020 [161]

Answer:

Goal is not achievable

Step-by-step explanation:

There are Two laps and the total distance here is 2miles. Now, it is said that he hopes to travel at a speed of 60mph, thus the total time that it would take him to complete the 2 miles distance race is 2/60 = 1/30 hours. Thus if he travels at a constant speed of 60mph he would finish the race at 1/30hours.

This means he will take a time of 1/60 hours per lap.

Now, he travels 30mph to complete 1 mile, the time taken for this is 1/30h

We now need to know the speed he travels on the second lap.

Average speed = Total distance/Total time

60mph = 2/( 1/30 + T2)

60 = 2/( (1 + 30T2)/30)

60 = 2 divided by (1 + 30T2)/30

60 = 2 * (30)/1 + 30T2

60( 1 + 30T2) = 60

1 + 30T2 = 1

30T2 = 0

T2 = 0

This means that he cannot achieve his goal again as he had taken the time meant for the whole race in a single lap.

4 0

4 years ago

Find the mass and the center of mass of a wire loop in the shape of a helix (measured in cm: x = t, y = 4 cos(t), z = 4 sin(t) f

Sholpan [36]

Answer:

Mass

$\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

Center of mass

Coordinate x

$\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate y

$\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Coordinate z

$\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

Step-by-step explanation:

Let W be the wire. We can consider W=(x(t),y(t),z(t)) as a path given by the parametric functions

x(t) = t

y(t) = 4 cos(t)

z(t) = 4 sin(t)

for 0 ≤ t ≤ 2π

If D(x,y,z) is the density of W at a given point (x,y,z), the mass m would be the curve integral along the path W

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt$

The density D(x,y,z) is given by

$D(x,y,z)=x^2+y^2+z^2=t^2+16cos^2(t)+16sin^2(t)=t^2+16$

on the other hand

$||W'(t)||=\sqrt{1^2+(-4sin(t))^2+(4cos(t))^2}=\sqrt{1+16}=\sqrt{17}$

and we have

$m=\displaystyle\int_{W}D(x,y,z)=\displaystyle\int_{0}^{2\pi}D(x(t),y(t),z(t))||W'(t)||dt=\\\\\sqrt{17}\displaystyle\int_{0}^{2\pi}(t^2+16)dt=\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)$

The center of mass is the point $(\bar x,\bar y,\bar z)$

where

$\bar x=\displaystyle\frac{1}{m}\displaystyle\int_{W}xD(x,y,z)\\\\\bar y=\displaystyle\frac{1}{m}\displaystyle\int_{W}yD(x,y,z)\\\\\bar z=\displaystyle\frac{1}{m}\displaystyle\int_{W}zD(x,y,z)$

We have

$\displaystyle\int_{W}xD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}t(t^2+16)dt=\\\\=\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)$

$\bar x=\displaystyle\frac{\sqrt{17}(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{(\displaystyle\frac{(2\pi)^4}{4}+32\pi)}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}yD(x,y,z)=\sqrt{17}\displaystyle\int_{0}^{2\pi}4cos(t)(t^2+16)dt=\\\\=16\sqrt{17}\pi$

$\bar y=\displaystyle\frac{16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

$\displaystyle\int_{W}zD(x,y,z)=4\sqrt{17}\displaystyle\int_{0}^{2\pi}sin(t)(t^2+16)dt=\\\\=-16\sqrt{17}\pi$

$\bar z=\displaystyle\frac{-16\sqrt{17}\pi}{\sqrt{17}(\displaystyle\frac{8\pi^3}{3}+32\pi)}=\displaystyle\frac{-16\pi}{(\displaystyle\frac{8\pi^3}{3}+32\pi)}$

3 0

3 years ago

Help ASAP I'm bad at solving this I will give out 20 points!

nexus9112 [7]

Answer:

(0,3/2), (0,2),(6,0),(9/4,0)

I hope it helpss..

3 0

4 years ago