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kirill [66]
3 years ago
7

There are 32 fowerds and 80 gards how many gards can go in the fowerds

Mathematics
2 answers:
scoray [572]3 years ago
6 0

Answer:

2.5

Step-by-step explanation:

2.5 fowerds can be put into 80 gards

Dafna1 [17]3 years ago
4 0
2.5 Fowerds can fit inside of 80 gards
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If ABCD is a parallelogram, what is the value of x?<br>​
Alexxandr [17]

x=44 since angle a and c  is congruent

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Need help please brainliest and 50 points
irina [24]

The slope is 2/-2, so -1. The y-intercept is 2. So, y=-x+2

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45.7 kilometers =<br> meters
VashaNatasha [74]

Answer:

45.7 kilometers = 45,700 meters

Explanation:

Hi!
Okay, so basically, in one kilometer, there are 1000 meters.
Using this problem, all you have to do is <u>multiply 45.7 by 1000</u>, which is 45,700.

Whatever value is used for kilometers (ex. 2), just multiply by 1000, and you’ll get the value as meters (ex. 2 x 1000 = 2000 meters).


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8 0
2 years ago
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Is 0.25 kilogram equivalent to any fraction
tester [92]
Equivilent means equals to
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4 0
3 years ago
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Where does the helix r(t) = cos(πt), sin(πt), t intersect the paraboloid z = x2 + y2? (x, y, z) = What is the angle of intersect
Colt1911 [192]

Answer:

Intersection at (-1, 0, 1).

Angle 0.6 radians

Step-by-step explanation:

The helix r(t) = (cos(πt), sin(πt), t) intersects the paraboloid  

z = x2 + y2 when the coordinates (x,y,z)=(cos(πt), sin(πt), t) of the helix satisfy the equation of the paraboloid. That is, when

\bf (cos(\pi t), sin(\pi t), t)

But  

\bf cos^2(\pi t)+sin^2(\pi t)=1

so, the helix intersects the paraboloid when t=1. This is the point

(cos(π), sin(π), 1) = (-1, 0, 1)

The angle of intersection between the helix and the paraboloid is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.

The <em>tangent vector</em> to the helix in t=1 is

r'(t) when t=1

r'(t) = (-πsin(πt), πcos(πt), 1), hence

r'(1) = (0, -π, 1)

A normal vector to the tangent plane of the surface  

\bf z=x^2+y^2

at the point (-1, 0, 1) is given by

\bf (\frac{\partial f}{\partial x}(-1,0),\frac{\partial f}{\partial y}(-1,0),-1)

where

\bf f(x,y)=x^2+y^2

since

\bf \frac{\partial f}{\partial x}=2x,\;\frac{\partial f}{\partial y}=2y

so, a normal vector to the tangent plane is

(-2,0,-1)

Hence, <em>a vector in the same direction as the projection of the helix's tangent vector (0, -π, 1) onto the tangent plane </em>is given by

\bf (0,-\pi,1)-((0,-\pi,1)\bullet(-2,0,-1))(-2,0,1)=(0,-\pi,1)-(-2,0,1)=(2,-\pi,0)

The angle between the tangent vector to the curve and the tangent plane to the paraboloid equals the angle between the tangent vector to the curve and the vector we just found.  

But we now

\bf (2,-\pi,0)\bullet(0,-\pi,1)=\parallel(2,-\pi,0)\parallel\parallel(0,-\pi,1)\parallel cos\theta

where  

\bf \theta= angle between the tangent vector and its projection onto the tangent plane. So

\bf \pi^2=(\sqrt{4+\pi^2}\sqrt{\pi^2+1})cos\theta\rightarrow cos\theta=\frac{\pi^2}{\sqrt{4+\pi^2}\sqrt{\pi^2+1}}=0.8038

and

\bf \theta=arccos(0.8038)=0.6371\;radians

7 0
3 years ago
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