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nata0808 [166]
1 year ago
15

a 20 cm nail just fits inside a cylindrical can. three identical spherical balls need to fit entirely within the can. what is th

e maximum radius of each ball?
Mathematics
1 answer:
DIA [1.3K]1 year ago
5 0

The maximum radius of each ball is 3.33 cm.

<h3>What is a Cylinder?</h3>

A cylinder is a three-dimensional figure which is made of two parallel circular bases separated at a certain distance by a curved surface of the same radius as that of the bases.

A 20 cm nail perfectly fits the cylinder.

So the height of the cylinder = 20cm.

A sphere is a three-dimensional geometrical figure analogous to a circle. All the points of the sphere are at the same distance from the center.

If 3 spheres have to be fitted inside the cylinder,

The sum of the diameter of the spheres will be equal to the height of the cylinder.

The spherical balls are identical and so will have the same diameter

Let the maximum diameter of the sphere that can fit in the cylinder is d cm

Then,

d + d + d = 20

3d = 20

d = 20/3 = 6.67 cm

r = 3.33 cm

The maximum radius that the spheres can have to fit in the cylinder is 3.33 cm.

To know more about Cylinder

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Graph the function f(x) = x3 – 5x2 + 6x using the graphing calculator. What are the solutions to the related equation? Check all
Juli2301 [7.4K]

the solutions to the related equation are 0,2,3 .

<u>Step-by-step explanation:</u>

Here we have , function  f(x) = x3 – 5x2 + 6x . Graph of this function is given below  . We need to find What are the solutions to the related equation . Let's find out:

Solution of graph means the  value of x at which the value of f(x) or function is zero . We can determine this by seeing the graph as at what value of x does the graph intersect or cut x-axis !

At x = 0 .

From the graph , at x=0 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 0^3- 5(0)^2 + 6(0)

⇒ f(0) =0

At x = 2 .

From the graph , at x=2 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 2^3- 5(2)^2 + 6(2)

⇒ f(0) =0

At x=3 .

From the graph , at x=3 we have f(x) = 0 i.e.

⇒ f(x) = x^3- 5x^2 + 6x

⇒ f(0) = 3^3- 5(3)^2 + 6(3)

⇒ f(0) =0

Therefore , the solutions to the related equation are 0,2,3 .

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A. Evaluate ∫20 tan 2x sec^2 2x dx using the substitution u = tan 2x.
irakobra [83]

Answer:

The integral is equal to 5\sec^2(2x)+C for an arbitrary constant C.

Step-by-step explanation:

a) If u=\tan(2x) then du=2\sec^2(2x)dx so the integral becomes \int 20\tan(2x)\sec^2(2x)dx=\int 10\tan(2x) (2\sec^2(2x))dx=\int 10udu=\frac{u^2}{2}+C=10(\int udu)=10(\frac{u^2}{2}+C)=5\tan^2(2x)+C. (the constant of integration is actually 5C, but this doesn't affect the result when taking derivatives, so we still denote it by C)

b) In this case u=\sec(2x) hence du=2\tan(2x)\sec(2x)dx. We rewrite the integral as \int 20\tan(2x)\sec^2(2x)dx=\int 10\sec(2x) (2\tan(2x)\sec(2x))dx=\int 10udu=5\frac{u^2}{2}+C=5\sec^2(2x)+C.

c) We use the trigonometric identity \tan(2x)^2+1=\sec(2x)^2 is part b). The value of the integral is 5\sec^2(2x)+C=5(\tan^2(2x)+1)+C=5\tan^2(2x)+5+C=5\tan^2(2x)+C. which coincides with part a)

Note that we just replaced 5+C by C. This is because we are asked for an indefinite integral. Each value of C defines a unique antiderivative, but we are not interested in specific values of C as this integral is the family of all antiderivatives. Part a) and b) don't coincide for specific values of C (they would if we were working with a definite integral), but they do represent the same family of functions.  

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3 years ago
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Answer:

-72

Step-by-step explanation:

-32 + (2-6)(10)

-32 + (-4)(10)

-32 + -40

-72

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