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Blababa [14]
2 years ago
10

2/3 feet is how many inches

Mathematics
1 answer:
leonid [27]2 years ago
3 0

Answer:

(2/3) foot =  8 inches

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I’m confused on this one
GenaCL600 [577]

The slope of CD is

m = \dfrac{6 - -4}{2 - 0} = 5

The slope of the perpendicular bisector is the negative reciprocal,

n = -1/m = - \dfrac 1 5

The perpendicular bisector passes through M, the midpoint of CD

M =( (0+2)/2, (-4 + 6)/2) = (1, -1)

So point slope form for the perpendicular bisector is

y - -1 = - \frac 1 5(x - 1)

Solving for y gives slope intercept form.

y = - \frac 1 5 x - \frac 4 5

Not sure how to type that without spaces; we didn't have online homework back then.

Answer: y=(-1/5)x-4/5

3 0
3 years ago
Draw a line representing the rise And a line representing the run of the mind state is slope of the line in simplest form
mamaluj [8]

Answer:

The slope is (1/5)

Step-by-step explanation:

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7 0
2 years ago
A pyramid and a cone are both 8 centimeters tall and have the same volume.
notka56 [123]

Answer:

D. The horizontal cross-sections of the prisms at the same height must have the same area.

Step-by-step explanation: I dont know what the guy above is saying but this is the right answer

4 0
2 years ago
Evaluate the surface integral. s x ds, s is the part of the plane 18x + 9y + z = 18 that lies in the first octant.
IceJOKER [234]
In the first octant, the given plane forms a triangle with vertices corresponding to the plane's intercepts along each axis.

(x,0,0)\implies 18x+9\cdot0+0=18\implies x=1
(0,y,0)\implies 18\cdot0+9y+0=18\implies y=2
(0,0,z)\implies 18\cdot0+9\cdot0+z=18\implies z=18

Now that we know the vertices of the surface \mathcal S, we can parameterize it by

\mathbf s(u,v)=\langle(1-u)(1-v),2u(1-v),18v\rangle

where 0\le u\le1 and 0\le v\le1. The surface element is

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=2\sqrt{406}(1-v)\,\mathrm du\,\mathrm dv

With respect to our parameterization, we have x(u,v)=(1-u)(1-v), so the surface integral is

\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{406}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)^2\,\mathrm dv\,\mathrm du=\frac{\sqrt{406}}3
5 0
3 years ago
Help please!!!!””””””””””
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