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slavikrds [6]
3 years ago
13

Which conclusion of the following statement must always be true?

Mathematics
1 answer:
erica [24]3 years ago
4 0

Answer:

4th answer

Step-by-step explanation:

angle 1 is congruent to angle 2, they stay the same..... as are corresponding angles that also appear when parallel lines are cut by a transversal .

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A line passes through the points (5, –8) and (6, 2). What is the slope?
Andreyy89

Answer:

use cymath it really helps with finding slopes

Step-by-step explanation:

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3 years ago
The row-reduced form of a system of three linear equations in three variables is given by this matrix. How many solutions does t
Damm [24]

Answer:

B. infinite solutions along a line

Step-by-step explanation:

The matrix is equivalent to the equations ...

x = 2

y = 9

0z = 0 . . . . . . true for any z

This tells you the solutions are ...

(x, y, z) = (2, 9, z)

This defines a line parallel to the z-axis.

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4 years ago
danni is working a sample booth at the national soda and beverage convention. she is paid $1.50 for each of the first ten tastin
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7 0
3 years ago
Read 2 more answers
The larger triangle is a dilation of the smaller triangle with a center of dilation at (1,−1).
statuscvo [17]

Answer:

just 4 I think

Step-by-step explanation:

the bottom of the bigger triangle is 8 and the little one is 2

8÷2= 4

3 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
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