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masya89 [10]
3 years ago
11

What happens when you square a cubed root or cube a square root and problems like these. Show examples.

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
3 0
Cube root of any value x means the power of or exponent of 1/3 of x.
X^1/3
if you square it,
the power of a power is multiplied according to exponent laws.
Hence,
===> (X^1/3)^2
         = X^2/3

Now,
Square root of any value x means the power of or exponent of 1/2 of x.
X^1/2
if you cube it,
the power of a power is multiplied according to exponent laws.
<span>Hence,
</span>===> <span>(X^1/2)^3
</span>         <span>= X^3/2
Hope it helps</span>
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A set of equations is given below
Alex_Xolod [135]

Answer:

No solution

Step-by-step explanation:

This is no solution due to the fact that the slope of both of the equation are the same. So they are parallel. If the b value is the same then it is infinitely many solutions. Due to the fact that the b value is not the same they are a set of parallel line. Parallel line do not touch or intersect ever so there for this is a no solution set.

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3 years ago
Due SOON!!
DedPeter [7]

Answer:

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Step-by-step explanation:

5/6x - 1/5x = 19

5(5/6x) - 6(1/5x) = 19

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19/30x=19

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2 years ago
A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in
Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

\frac{dr}{dt} =\frac{1}{2} --- rate of the radius

\frac{dh}{dt} =-\frac{1}{3} --- rate of the height

r = 2

h = \frac{1}{3}

Required

Determine the rate of change of the cone volume

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V = \frac{\pi}{3}r^2h

Differentiate with respect to time (t)

\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})

Substitute values for the known variables

\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)

\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}

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\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}

\frac{dV}{dt} = -\frac{88}{63}

\frac{dV}{dt} =-1.396in^3/min

The volume is decreasing at the rate of 1.396 cubic inches per minute

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