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Alik [6]
3 years ago
14

One more.... even answering one helps me a lot thanks :) :) :)

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
7 0
1)
2y = -x + 6
x + 2y = 6 ...this is standard form

2)
9x + 2y = 6
2y = -9x + 6
  y = -4.5x + 3 .....this is slope intercept form

3)
y - 3 = 4(x + 1)
y - 3 = 4x + 4
4x - y = -7  .......this is standard form

4)

y + 7 = -2(x - 3)
y + 7 = -2x + 6
y = -2x - 1 .........this is slope intercept form

hope it helps
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Mr. Santiago can buy light fixtures in packages of 12 and light bulbs in packages of 9. He bought the fewest number of light fix
liq [111]

Answer:

  • 3 and 4 packages

Step-by-step explanation:

We need to find the LCM of 12 and 9.

  • 9 = 3*3
  • 12 = 2*2*3

<u>Their LCM is:</u>

  • LCM(12, 9) = 2*2*3*3 = 36

Number of light fixtures and light bulbs is 36.

<u>Number of packages will be:</u>

  • 36/12 = 3 and
  • 36/9 = 4 respectively.
6 0
2 years ago
Read 2 more answers
5) Which graph shows the solution of the system of inequalities 2x - y&lt;-6 and y &lt; 2x - 3?
liq [111]

Answer:

B

Step-by-step explanation:

4 0
2 years ago
Area of regular polygons
Finger [1]
Hello!

To find the area of a hexagon you do 3\frac{3 \sqrt{3} }{2} } a^{2} where a is one of the sides

Since the perimeter is 60 we can do 60/ the sides of the shape

60/6 = 10

So one side is equal to 10

you put that into the formula to get the area of a hexagon and you get 259.81

Hope this helps!
8 0
3 years ago
Birthstones of students in your class<br> qualitative Or quantitative??
Alik [6]
Quantitive since it can be counted
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3 years ago
A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (&gt; r) from th
jeyben [28]

Consider a circle with radius r centered at some point (R+r,0) on the x-axis. This circle has equation

(x-(R+r))^2+y^2=r^2

Revolve the region bounded by this circle across the y-axis to get a torus. Using the shell method, the volume of the resulting torus is

\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx

where 2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}.

So the volume is

\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx

Substitute

x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt

and the integral becomes

\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt

Notice that \sin t\cos^2t is an odd function, so the integral over \left[-\frac\pi2,\frac\pi2\right] is 0. This leaves us with

\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt

Write

\cos^2t=\dfrac{1+\cos(2t)}2

so the volume is

\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}

6 0
3 years ago
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