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3 years ago

Find the volume of a right circular cone that has a height of 16 m and a base with a

Mathematics

1 answer:

anzhelika [568]3 years ago

8 0

Answer:

The volume of a right circular cone that has a height of 16 m and a base with a circumference of 14.7 m is 3,620.62 m³

Step-by-step explanation:

The right cone (or cone of revolution, or right circular cone) is the solid of revolution formed by rotating a right triangle around one of its legs. The bottom circle of the cone is called the base. That is, a cone is a three-dimensional figure with a circular base. A curved surface connects the base and the vertex.

The volume of a 3-dimensional solid is the amount of space it occupies.

The volume V of a cone with radius r is one-third the area of the base B times the height h. This is:

$V=\frac{1}{3}*A*h$

where A=π*r²

Then: V= $\frac{1}{3}$ *π*r²*h

In this case r=14.7 m and h=16 m. Replacing:

V= $\frac{1}{3}$ *π*(14.7 m)²*16 m

Solving, you get:

V=3,620.62 m³

<u><em>The volume of a right circular cone that has a height of 16 m and a base with a circumference of 14.7 m is 3,620.62 m³</em></u>

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12. The point P lies on a line through the midpoint of AB and perpendicular to AB. If AP=15 in., what is the length PB? How do y

Mekhanik [1.2K]

Answer:

PB = 15

Step-by-step explanation:

A midpoint is a point in the center of a line. Since P is the midpoint of AB, the length of AP = PB. Therefore, if AP = 15, PB = 15.

8 0

3 years ago

12(3-2x)=84 solve for x.

OlgaM077 [116]

12(3-2x) = 84

do distributive multiplication:

12(3) + 12(-2x) = 84
36 - 24x = 84

Deduct 36 to both sides

36 - 36 - 24x = 84 - 36
-24x = 48

divide -24 to both sides

-24x / -24 = 48/-24
x = -2

To check:
12(3-2x) = 84
12(3-2(-2) = 84
12(3+4) = 84
12(7) = 84
84 = 84

7 0

4 years ago

Read 2 more answers

Can someone thoroughly explain this implicit differentiation with a trig function. No matter how many times I try to solve this,

Anton [14]

Answer:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

Step-by-step explanation:

So we have the equation:

$\tan(x-y)=\frac{y}{8+x^2}$

And we want to find dy/dx.

So, let's take the derivative of both sides:

$\frac{d}{dx}[\tan(x-y)]=\frac{d}{dx}[\frac{y}{8+x^2}]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[\tan(x-y)]$

We can use the chain rule, where:

$(u(v(x))'=u'(v(x))\cdot v'(x)$

Let u(x) be tan(x). Then v(x) is (x-y). Remember that d/dx(tan(x)) is sec²(x). So:

$=\sec^2(x-y)\cdot (\frac{d}{dx}[x-y])$

Differentiate x like normally. Implicitly differentiate for y. This yields:

$=\sec^2(x-y)(1-y')$

Distribute:

$=\sec^2(x-y)-y'\sec^2(x-y)$

And that is our left side.

Right Side:

We have:

$\frac{d}{dx}[\frac{y}{8+x^2}]$

We can use the quotient rule, where:

$\frac{d}{dx}[f/g]=\frac{f'g-fg'}{g^2}$

f is y. g is (8+x²). So:

$=\frac{\frac{d}{dx}[y](8+x^2)-(y)\frac{d}{dx}(8+x^2)}{(8+x^2)^2}$

Differentiate:

$=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

And that is our right side.

So, our entire equation is:

$\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}$

To find dy/dx, we have to solve for y'. Let's multiply both sides by the denominator on the right. So:

$((8+x^2)^2)\sec^2(x-y)-y'\sec^2(x-y)=\frac{y'(8+x^2)-2xy}{(8+x^2)^2}((8+x^2)^2)$

The right side cancels. Let's distribute the left:

$\sec^2(x-y)(8+x^2)^2-y'\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy$

Now, let's move all the y'-terms to one side. Add our second term from our left equation to the right. So:

$\sec^2(x-y)(8+x^2)^2=y'(8+x^2)-2xy+y'\sec^2(x-y)(8+x^2)^2$

Move -2xy to the left. So:

$\sec^2(x-y)(8+x^2)^2+2xy=y'(8+x^2)+y'\sec^2(x-y)(8+x^2)^2$

Factor out a y' from the right:

$\sec^2(x-y)(8+x^2)^2+2xy=y'((8+x^2)+\sec^2(x-y)(8+x^2)^2)$

Divide. Therefore, dy/dx is:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)+\sec^2(x-y)(8+x^2)^2}$

We can factor out a (8+x²) from the denominator. So:

$\frac{dy}{dx}=y'=\frac{\sec^2(x-y)(8+x^2)^2+2xy}{(8+x^2)(1+\sec^2(x-y)(8+x^2))}$

And we're done!

8 0

3 years ago

consider the equation a(4-x) =-3x + b What values of a and b make the solution of the equation x = -5

-BARSIC- [3]

Answer:

Step-by-step explanation:

hello :

put x = -5 in this equation : a(4-x) =-3x + b

a(4+5) =15 + b

9a = 15 +b

b = 9a - 15

the values of a and b is all order pair : ( a ; 9a-15) a real number

3 0

3 years ago

Without graphing, determine whether the following pair of lines are parallel, perpendicular, or neither parallel nor perpendicul

Tomtit [17]

Write the equation of the lines in slope-intercept form (y=mx+b)

First equation: It is already in slope-intercept form

$y=\frac{2}{5}x-4$

Second equation: solve y:

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Identify the slope of each line:

If two or more lines have the same slope then the lines are parallel

If two lines have slopes that are negative reciprocals then the lines are perpendicular

The two given lines have the same slope: 2/5. Then, they are parallel lines

4 0

1 year ago