X=12, -9 I hope this somewhat helps you :)
Answer:
![\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E4%7D)
Step-by-step explanation:
Maybe you want 2^(4/5) in radical form.
The denominator of the fractional power is the index of the root. Either the inside or the outside can be raised to the power of the numerator.
![2^{\frac{4}{5}}=\boxed{\sqrt[5]{2^4}=(\sqrt[5]{2})^4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%3D%5Cboxed%7B%5Csqrt%5B5%5D%7B2%5E4%7D%3D%28%5Csqrt%5B5%5D%7B2%7D%29%5E4%7D)
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In many cases, it is preferred to keep the power inside the radical symbol.
The sum of the first 75 terms of the arithmetic sequence that has 10th term as 16 and the 35th term as 66 is 5400.
<h3>How to find the sum of terms using Arithmetic sequence formula</h3>
aₙ = a + (n - 1)d
where
Therefore, let's find a and d
a₁₀ = a + (10 - 1)d
a₃₅ = a + (35 - 1)d
Hence,
16 = a + 9d
66 = a + 34d
25d = 50
d = 50 / 25
d = 2
16 - 9(2) = a
a = 16 - 18
a = -2
Therefore, let's find the sum of 75 terms of the arithmetic sequence
Sₙ = n / 2 (2a + (n - 1)d)
S₇₅ = 75 / 2 (2(-2) + (75 - 1)2)
S₇₅ = 37.5 (-4 + 148)
S₇₅ = 37.5(144)
S₇₅ = 5400
learn more on arithmetic sequence here: brainly.com/question/1687271
Answer:
48
240 ÷ 5 = 48
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