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Fantom [35]
3 years ago
10

Please help me with this i really need help

Mathematics
1 answer:
Lena [83]3 years ago
4 0

Answer:

Add all 25-29 ages together to get your answer

Step-by-step explanation:

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Find the sum of 2x2 3x - 4, 8 - 3x, and -5x2 2. -3x2 - 2 -7x2 - 2 -3x2 6 -7x2 - 6x - 2
Law Incorporation [45]
-3x to the second power - 2

I believe this is your answer. 
4 0
3 years ago
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
If y varies directly with x, and y= 10 when x = 4, find y when x = 10.​
mixer [17]

Answer:

17

Step-by-step explanation:

bc

6 0
3 years ago
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