Help me with this question

You are designing a container in the shape of a cylinder. The radius is 6 inches. You want the container to hold at least 324π c

goldfiish [28.3K]

Answer: 9 inches

Explaination: The volume of a cylinder, V = (πr^2)H, so H = V/(πr^2).

Using the parameters you specified, H = 324π/[(π)(6^2)] = 324π/36π = 324/36 = 9 inches

4 0

3 years ago

A candle is in the shape of a square pyramid. Its height is 5 cm and its slant height is 8 cm.

raketka [301]

Answer:

it is 5.9 A

Step-by-step explanation:

7 0

3 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

3 years ago

Mr. Diaz is building a fence around his yard. For numbers 3a-3b, choose the values and term that correctly describe the shape of

Inga [223]

3a) The figure has 5 sides and 5 vertices

3b) None of the sides are congruent, so the figure is not are guitar polygon

3 0

3 years ago

Prove this trigonometric equation;<br><br> - tan^2x + sec^2x = 1,

alekssr [168]

Hey there :)

- tan²x + sec²x = 1 or 1 + tan²x = sec²x

sin²x + cos²x = 1
Divide the whole by cos²x

$\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}$

$\frac{sinx}{cosx} = tanx$ so $\frac{sin^2x}{cos^2x} = tan^2x$
and
$\frac{1}{cosx} = secx$ so $\frac{1}{cos^2x} = sec^2x$

Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}

1 = - tan²x = sec²x or sec²x - tanx = 1

3 0

3 years ago