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Darina [25.2K]
2 years ago
13

PLEASE HELPP WHAT IS THE SIMPLEST FORM OF 3 square root of x to the 10

Mathematics
2 answers:
Nina [5.8K]2 years ago
4 0

Answer:

x^3 3 square root x

Step-by-step explanation:

lutik1710 [3]2 years ago
3 0

<span>A radical is a mathematical symbol used to represent the root of a number. Here’s a quick example: the phrase “the square root of 81” is represented by the radical expression . (In the case of square roots, this expression is commonly shortened to —notice the absence of the small “2.”) When we find  we are finding the non-negative number r such that , which is 9.</span>

 

<span>While square roots are probably the most common radical, we can also find the third root, the fifth root, the 10th root, or really any other nth root of a number. The nth root of a number can be represented by the radical expression.</span>

 

Radicals and exponents are inverse operations. For example, we know that 92 = 81 and  = 9. This property can be generalized to all radicals and exponents as well: for any number, x, raised to an exponent n to produce the number y, the nth root of y is x.

 

We can represent this property like this: . A warning though: it is always true if x ≥ 0, and it is always true if n is odd. But it is not true when x < 0 and n is even.

 

Why is this the case? It is because raising any number, positive or negative, to an even power has the effect of making the new number positive. This is not the case for odd exponents. For example, think about inserting x = -3 and n = 2 into the formula above.

 

<span>The radical would be written as , which works out to, or 3. But our initial x value was -3, so we are left with the statement 3 = -3. </span>

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Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

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2 years ago
The point g (2,2) is translated 3units to the right and 4units up what are the coordinates of this resulting point, g?
Svetlanka [38]

Answer:

(5,6)

Step-by-step explanation:

<em>Generally, is the coordinate (x,y) is translated right by a units and up by b units, the resulting coordinate wil be;</em>

g(X, Y)  = (x+a, y+b)

To the right is along the positive x axis while up is along the positive y axis

Given a = 3 and b = 4

g(X,Y) = (2 + 3, 2 + 4)

g(X, Y) = (5, 6)

Hence the resulting coordinate of g is (5,6)

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