Answer:
![2^{\frac{4}{5}}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D)
Step-by-step explanation:
Factor 16 using prime factorisation:
16 ÷ 2 = 8
8 ÷ 2 = 4
4 ÷ 2 = 2
⇒ 16 = 2⁴
Substitute 16 for 2⁴:
![\implies (\sqrt[5]{16})^1=(\sqrt[5]{2^4})^1](https://tex.z-dn.net/?f=%5Cimplies%20%28%5Csqrt%5B5%5D%7B16%7D%29%5E1%3D%28%5Csqrt%5B5%5D%7B2%5E4%7D%29%5E1)
![\textsf{Apply exponent rule}\quad \sqrt[n]{a^b} =a^{\frac{b}{n}}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5Eb%7D%20%3Da%5E%7B%5Cfrac%7Bb%7D%7Bn%7D%7D%3A)
![\implies (\sqrt[5]{2^4})^1=(2^{\frac{4}{5}})^1](https://tex.z-dn.net/?f=%5Cimplies%20%28%5Csqrt%5B5%5D%7B2%5E4%7D%29%5E1%3D%282%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%29%5E1)
![\textsf{Apply exponent rule} \quad (a^b)^c=a^{bc}:](https://tex.z-dn.net/?f=%5Ctextsf%7BApply%20exponent%20rule%7D%20%5Cquad%20%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D%3A)
![\implies (2^{\frac{4}{5}})^1=2^{\frac{4}{5}}](https://tex.z-dn.net/?f=%5Cimplies%20%282%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%29%5E1%3D2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D)
Answer:
<h2>What is ASAP </h2>
Step-by-step explanation:
<h2>WHAT IS ASAP </h2>
You are given a vector in the XY plane that has a magnitude of 87. 0 units and a y component of -66. 0 units. The direction of the vector V is;
<h3>How to know the direction of a vector?</h3>
We know that the formula for 2 vectors like this in the x and y directions is; A = xi^ + yj^
Where A is the magnitude of the resultant
x is the value of the x-component
y is the value of the y-component
We are given;
Magnitude of vector = 84 units
Y-component of the vector = -67 units
Thus,
![A = \sqrt(x^2 + y^2)\\\\87 = \sqrt(x^2+ (-66)^2)\\\\87^2= x^2+ 4356\\\\7569 = x^2+ 4356\\\\x= \sqrt(7569 - 4356)\\\\x = 56.68 units](https://tex.z-dn.net/?f=A%20%3D%20%5Csqrt%28x%5E2%20%2B%20y%5E2%29%5C%5C%5C%5C87%20%3D%20%5Csqrt%28x%5E2%2B%20%28-66%29%5E2%29%5C%5C%5C%5C87%5E2%3D%20x%5E2%2B%204356%5C%5C%5C%5C7569%20%3D%20x%5E2%2B%204356%5C%5C%5C%5Cx%3D%20%5Csqrt%287569%20-%204356%29%5C%5C%5C%5Cx%20%3D%2056.68%20units)
From A above, let us take the positive value of the x-component and as such our original vector will be;
A = 56.68i^ - 66j^
The direction of the vector V is;
![\theta = tan^{-1} \dfrac{y}{x} \\\\\theta = tan^{-1} \dfrac{-66}{56.68} \\\theta =-27.15°](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5Cdfrac%7By%7D%7Bx%7D%20%20%5C%5C%5C%5C%5Ctheta%20%3D%20%20tan%5E%7B-1%7D%20%5Cdfrac%7B-66%7D%7B56.68%7D%20%20%5C%5C%5Ctheta%20%3D-27.15%C2%B0)
Since it points entirely to the negative x-axis, then the angle is;
180 - (-27.15) = 207.15°
Learn more about vectors;
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