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OLEGan [10]
3 years ago
12

Volume of cylinder in terms of pi

Mathematics
1 answer:
amid [387]3 years ago
6 0

The formula to calculate the volume of a cylinder is: V = Π h r², that is, Pi per height per radius squared Steps to follow: one When calculating the volume of a cylinder, you will need to measure the height (h) of the cylinder, a value that sometimes also receives the name of length. Let's give a concrete example to be able to apply the formula and that way you understand better how to carry out the operations. In this way, we will assume that the height or length of the cylinder is 10 cm. two On the other hand, you must also have the measure of the radius (r) of the cylinder, that is, the distance from the outer edge to the center of the circle that serves as the base of this polyhedral figure. Make sure you use the same unit of measure for each dimension, if they are expressed in different units, you must do the conversion to obtain the equivalence. Following the example, let's give the radio a value of 3 cm 3 So, knowing that the formula to calculate the volume of a cylinder is: V = Π h r² We just need to know how much pi is worth, remember? Yes, we can round it to 3.14 and, therefore, we will only need to substitute each value in the formula. Next, you must calculate the square of the radius (r²), multiplying the value by itself and then multiply the result of the height by Pi, by the result of the radius squared. V = 3.14 x 10 x 3² = 3.14 x 10 x 9 = 282.6 cm³ Remember to write the answer in the appropriate cubic unit of measure. So you can understand where you get the formula to calculate the volume of a cylinder: If you disarm a cylinder in your hands you will have two equal circles and a rectangle, right? First you must use the formula to take out the area of ​​a circle that is: A = (Π) (r²) When you get the area of ​​the circle you multiply it by the height of the cylinder and the result is the volume of the cylinder. Always remember to be careful with the units of measurement you use in calculations.

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How can i find the midpoint of a line segment ?​
eimsori [14]

Answer:

add both x coordinates and divide them by 2.

add both y coordinates and divide them by 2.

Now final product should be (x,y)

Step-by-step explanation:

Example.

let's take 2 points:

(2,5) and (7, 9)

let's add both x coordinates.

2+7 = 9

now add both y coordinates.

5+9 = 14

Now divide both by 2.

Final answer should be (4.5, 7) = this is your midpoint

8 0
3 years ago
Simplify i^82 pls...thanks :)
Fofino [41]
To simplify i^82, divide 82 by. the answer is 20 with remainder 2
8 0
3 years ago
Read 2 more answers
In a 50-gram portion of cereal, 8 grams of it are sugar . Which of the following represents the percent of the 50-grams that is
andre [41]

Answer:

3) 16

8 is 16% of 50

4 0
3 years ago
Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

Answer:

X(16)=25.71grams

Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

\frac{2}{3}Xg of A and \frac{1}{3}Xg of B

Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

5 0
3 years ago
In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

8 0
3 years ago
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