Ask question

Ask question

All categories

3 years ago

11

Describe different ways in which a plane might the cylinder,and the cross section the results.

1 answer:

Mademuasel [1]3 years ago

5 0

The plane could be:
-perpendicular to the axis --> gives a circle
-parallel to the axis and tangent to the cylinder --> gives a line
-parallel to the axis and not tangent to the cylinder --> gives a pair of parallel lines
-neither perpendicular nor parallel to the axis --> gives an ellipse

You might be interested in

Triangle XYZ is located at X (−2, 1), Y (−4, −3), and Z (0, −2). The triangle is then transformed using the rule (x−1, y+3) to f

Airida [17]

<span>You have to apply the rule to the XYZ coordinates, so X'= (-2-1, 1+3) Y'= (-4-1, -3+3) Z'= (0-1, -2+3) So the new coordinates are X' (-3, 4), Y' (-5, 0), and Z'(-1, 1) This is actually translation not transformation, as the shape doesn't change it just moves</span>

6 0

3 years ago

If possible please answer

mrs_skeptik [129]

Answer:

<h2>-3<x≤1 -or- (-3,1]</h2>

4 0

3 years ago

Pls help LOL I need helo

Andrews [41]

500$ if that’s an answer

5 0

2 years ago

Amy wants to carpet a room that is 12 feet by 8 feet.How many square yards of carpet will she need to complete the room?

Ede4ka [16]

96 square yards would complete the room

7 0

3 years ago

Calculate two iterations of Newton's Method for the function using the given initial guess. (Round your answers to four decimal

pickupchik [31]

Answer:

The first and second iteration of Newton's Method are 3 and $\frac{11}{6}$ .

Step-by-step explanation:

The Newton's Method is a multi-step numerical method for continuous diffentiable function of the form $f(x) = 0$ based on the following formula:

$x_{i+1} = x_{i} -\frac{f(x_{i})}{f'(x_{i})}$

Where:

$x_{i}$ - i-th Approximation, dimensionless.

$x_{i+1}$ - (i+1)-th Approximation, dimensionless.

$f(x_{i})$ - Function evaluated at i-th Approximation, dimensionless.

$f'(x_{i})$ - First derivative evaluated at (i+1)-th Approximation, dimensionless.

Let be $f(x) = x^{2}-8$ and $f'(x) = 2\cdot x$ , the resultant expression is:

$x_{i+1} = x_{i} -\frac{x_{i}^{2}-8}{2\cdot x_{i}}$

First iteration: ( $x_{1} = 2$ )

$x_{2} = 2-\frac{2^{2}-8}{2\cdot (2)}$

$x_{2} = 2 + \frac{4}{4}$

$x_{2} = 3$

Second iteration: ( $x_{2} = 3$ )

$x_{3} = 3-\frac{3^{2}-8}{2\cdot (3)}$

$x_{3} = 2 - \frac{1}{6}$

$x_{3} = \frac{11}{6}$

7 0

3 years ago