Answer: D) cube root of 16
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Explanation:
The rule we use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
In this case, x = 4, m = 2 and n = 3.
So,
![x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E2%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B16%7D%5C%5C%5C%5C%5C%5C)
Showing that the original expression turns into the cube root of 16.
It is 13,14 because it goes up two then one one then two again which will then mess up the rotation
Answer:
TU ≈ 12.96
Step-by-step explanation:
Using the Altitude on Hypotenuse theorem
(leg of outer triangle)² = (part of hypotenuse below it) × (whole hypotenuse)
TU² = UV × SU = 6 × 28 = 168 ( take square root of both sides )
TU =
≈ 12.96 ( to the nearest hundredth )
Answer:
lw +
× π ×
⇒ Answer D is correct
Step-by-step explanation:
First, let us find the area of the semi-circle.
Area =
× π × r²
<u>Given that,</u>
diameter of the semi-circle is ⇒ <em>l</em>
∴ radius ⇒ <em>l / 2</em>
<u>Let us find it now.</u>
Area =
× π × r²
Area =
× π × 
<u> </u>
Secondly, let us find the area of the rectangle.
Area = length × width
<u>Given that,</u>
length ⇒ <em>l</em>
width ⇒ w
<u>Let us find it now.</u>
Area = length × width
Area = l ×w
Area = lw
<u> </u>
And now let us <u>find the total area.</u>
Total area = Area of the rectangle + Area of the semi - circle
Tota area = lw +
× π × 