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Bond [772]
3 years ago
14

How to right a radical in exponential form

Mathematics
1 answer:
Law Incorporation [45]3 years ago
3 0

Answer:

  \sqrt[n]{x}=x^{\frac{1}{n}}

Step-by-step explanation:

The index of a radical is the denominator of a fractional exponent, and vice versa. If you think about the rules of exponents, you know this must be so.

For example, consider the cube root:

\sqrt[3]{x}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}=(\sqrt[3]{x})^3=x\\\\(x^{\frac{1}{3}})^3=x^{\frac{3}{3}}=x^1=x

That is ...

\sqrt[3]{x}=x^{\frac{1}{3}} \quad\text{radical index = fraction denominator}

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Per single gallon of gas, Gina's vehicle can go 16 more miles than Amanda's vehicle. If the combined distance the vehicles can t
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A birthday celebration meal is $46.80 including​ tax, but not the tip. find the total cost if a 10% tip is added to the cost of
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Running times for 400 meters are Normally distributed for young men between 18 and 30 years of age with a mean of 93 seconds and
Luda [366]

Answer:

C. 45 and 141 seconds

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 93 seconds

Standard deviation = 16 seconds

99.7% of running times are approximately between:

By the Empirical rule, within 3 standard deviations of the mean, so between 3 standard deviations below the mean and 3 standard deviations above the mean

3 stnadard deviations below the mean

93 - 3*16 = 45 seconds

3 standard deviations above the mean

93 + 3*16 = 141 seconds

The correct answer is:

C. 45 and 141 seconds

3 0
3 years ago
Use​ Newton's method to find an approximate solution of ln(x)=10-x. Start with x_0 =9 and find x_2 .
vova2212 [387]

Answer:

x₂ = 7.9156

Step-by-step explanation:

Given the function  ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ -  f(xₙ)/f'(xₙ)

If f(x) = ln(x)+x-10

f'(x) = 1/x + 1

f(9) = ln9+9-10

f(9) = ln9- 1

f(9) = 2.1972 - 1

f(9) = 1.1972

f'(9) = 1/9 + 1

f'(9) = 10/9

f'(9) = 1.1111

x₁ = x₀ -  f(x₀)/f'(x₀)

x₁ = 9 -  1.1972/1.1111

x₁  = 9 - 1.0775

x₁  = 7.9225

x₂ = x₁ -  f(x₁)/f'(x₁)

x₂ = 7.9225 -  f(7.9225)/f'(7.9225)

f(7.9225) = ln7.9225 + 7.9225 -10

f(7.9225) = 2.0697 + 7.9225 -10

f(7.9225) = 0.0078

f'(7.9225) = 1/7.9225 + 1

f'(7.9225) = 0.1262+1

f'(7.9225) = 1.1262

x₂ = 7.9225 - 0.0078/1.1262

x₂ = 7.9225 - 0.006926

x₂ = 7.9156

<em>Hence the approximate value of x₂ is 7.9156</em>

7 0
3 years ago
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