What is the length of a rectangle forms from the lateral surface of a cylinder?

2 answers:

8 0

So the lenght of a rectangle forms from the lateral surface of a cylinder will be the circumference of the base of this cylinder so what is choice B. so 2pi*r

hope helped

mihalych1998 [28]3 years ago

6 0

The length of a rectangle that forms from a lateral surface of a cylinder is C. 2 Pi rh.

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Help please this is for my homework that is due to tomorrow!!

Murljashka [212]

The given polygons are similar

Step-by-step explanation:

By observing the given diagrams, we can see that the respective angles are congruent.

Two shapes are said to be similar if they have same shapes or one shape is created by uniformly increasing or decreasing the side of other.

For this purpose we find the scale factor of greater to smaller image, if the scale factor is same for each side then the two shapes are similar

So,

The scale factor is:

$\frac{EF}{AB} = \frac{3}{1} = 3\\\\\frac{FG}{BC} = \frac{9}{3} = 3\\\\\frac{GH}{CD} = \frac{15}{5} = 3\\\\\frac{EH}{AD} = \frac{21}{7} = 3$

We can see that the scale factor is same for all sides.

Hence,

The given polygons are similar

Keywords: Polygons, similarity

Learn more about polygons at:

#LearnwithBrainly

5 0

3 years ago

Given the function f(x)=x+5-1 choose the correct domain

ElenaW [278]

Answer:

The domain of the function is all real numbers.

Step-by-step explanation:

I graphed the function on the graph below.

If this answer is correct, please make me Brainliest!

6 0

3 years ago

Assume that f is continuous on [-4,4] and differentiable on (-4,4). The table gives some values of f'(x) x: -4, -3, -2, -1, 0, 1

kondaur [170]

$f$ will be increasing on the intervals where $f'(x)>0$ and decreasing wherever $f'(x)$ . Local extrema occur when $f'(x)=0$ and the sign of $f'(x)$ changes to either side of that point.

$f'(x)$ is positive when $x$ is between -4 and some number between -2 and -1, and also 2 (exclusive) and 4, so you can estimate that $f(x)$ is increasing on the intervals [-4, -2] and (2, 4].

$f'(x)$ is negative when $x$ is between some number between -2 and -1, up to some number less than 2. So $f(x)$ is decreasing on the interval [-1, 1].

You then have two possible cases for extrema occurring. The sign of $f'(x)$ changes for some $x$ between -2 and -1, and again to either side of $x=2$ .