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Fynjy0 [20]
2 years ago
8

How do you make 8.25 a whole?​

Mathematics
1 answer:
vampirchik [111]2 years ago
8 0
You can round it to 9.
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In this fulcrum, the weights are perfectly balanced. How far must the fulcrum be located from the 40 lb. weight if the bar is 9
creativ13 [48]
So...we can write eqn as

40x = 50(x-1)
so. -10x=-50
so. x=5

therefore .... the distance of 40 lb will be 5 ft ...answer !!!
4 0
3 years ago
Read 2 more answers
Please hellppp i will mark brainliest ​
castortr0y [4]
30 degrees it's right
8 0
3 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:

<u>Mass</u>

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

<u>Center of mass</u>

<em>Coordinate x</em>

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

<em>Coordinate y</em>

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

<em>Coordinate z</em>

\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)

so

\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
A student creates a function, h(x), to model the path of a T-shirt
Radda [10]

HELP ME PLES

PERIOD, FREQUENCY OR AMPLITUDE

1. Doesn't change period

2. More of this means more energy

3. Increases as a pendulum swings back and forth faster

4. Measured in cycles per second

5. Measured in meters or centimeters

6. This is decreases with smaller swing

7. If the frequency increases, this decreases

8. Measured in Hertz

9. Measured in seconds

10. if it swings back and forth slower, this decrease

11. As it dampens, this decreases

5 0
3 years ago
Anybody know the answer to any of these
Tresset [83]
<h3>Answers :-</h3>
  • Answer 1 → tanZ = 21/28 = 3/4
  • Answer 2 → cosC = 16/34 = 8/17
  • Answer 3 → sinC = 28/35 = 4/5
  • Answer 4 → tanX = 24/32 = 3/4
  • Answer 5 → cosA = 20/34 = 10/17
  • Answer 6 → sinA = 32/40 = 4/5
  • Answer 7 → sinZ = 24/32 = 3/4
  • Answer 8 → sinC = 14/50 = 7/25
7 0
3 years ago
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