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Margaret [11]
1 year ago
9

Create the smallest cone possible with the tool, and record the values of the radius, height, and volume in terms of . Then scal

e the original CONE
by the given scale factors, and record the resulting volumes (in terms of 7), to verify that the formula V=V xk³ holds true for a cone.
Mathematics
1 answer:
Nata [24]1 year ago
8 0

The Created cone  with the possible record  of the the values of the radius, and  height is given in the image attached.

<h3>How do yo create the volume of the smallest cone?</h3>

A scale factor is known to be a number that is known to often multiplies (doubles or triples, etc.,“scales”) a given quantity.

Since the volume is not attached, it will be manually created here.

Note that the resulting volumes must be in terms of 7.

Hence:

Volume  Formula:   V=V  x  k³

When scale factor is 2, radius  = 4, height  = 8,  then volume will be:

7π x 2³

=  56π

When scale factor is 3, radius  = 6, height  = 18,  then volume will be:

7π x 3³

= 189π

When scale factor is 4, radius  = 8, height  = 24,  then volume will be:

7π x 4³

= 448π

Hence the volume that will be fixed into the scale factor diagram will be:

  • 56π
  • 189π
  • 448π

Learn more about scale factor from

brainly.com/question/17093741

#SPJ1

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How many extraneous solutions does the equation below have? StartFraction 9 Over n squared 1 EndFraction = StartFraction n 3 Ove
BigorU [14]

The equation has one extraneous solution which is n ≈ 2.38450287.

Given that,

The equation;

\dfrac{9}{n^2+1} =\dfrac{n+3}{4}

We have to find,

How many extraneous solutions does the equation?

According to the question,

An extraneous solution is a solution value of the variable in the equations, that is found by solving the given equation algebraically but it is not a solution of the given equation.

To solve the equation cross multiplication process is applied following all the steps given below.

\rm \dfrac{9}{n^2+1} =\dfrac{n+3}{4}\\\\9 (4) = (n+3) (n^2+1)\\\\36 = n(n^2+1) + 3 (n^2+1)\\\\36 = n^3+ n + 3n^2+3\\\\n^3+ n + 3n^2+3 - 36=0\\\\n^3+ 3n^2+n -33=0\\

The roots (zeros) are the  x  values where the graph intersects the x-axis. To find the roots (zeros), replace  y

with  0  and solve for  x. The graph of the equation is attached.

n  ≈  2.38450287

Hence, The equation has one extraneous solution which is n  ≈  2.38450287

For more information refer to the link.

brainly.com/question/15070282

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