C-26.5 is the answer I think hope it helped
The domain of the function is the set of all real numbers and the range of the function is the set of all values greater than -2
<h3>How to determine the domain and the range?</h3>
The function is given as:
f(x) = 2(x -4)^2 - 2
A quadratic function can take any real number as its input.
So, the domain of the function is the set of all real numbers
The vertex of the above function is:
Vertex = (4, -2)
And the leading coefficient is:
a = 2
The y value of the vertex is;
y = -2
Because the value of a is positive, then the vertex is a minimum.
This means that the range of the function is the set of all values greater than -2
Read more about domain and range at:
brainly.com/question/10197594
#SPJ1
Answer:
We conclude that:
![\:\sqrt[3]{200k^{15}}=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%5C%3A%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Hence, option B is correct.
Step-by-step explanation:
Given the expression
![\sqrt[3]{200k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D)
Apply radical rule:
![\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200%2C%5C%3Ab%5Cge%200)
so the expression becomes
![\sqrt[3]{200k^{15}}=\sqrt[3]{200}\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200k%5E%7B15%7D%7D%3D%5Csqrt%5B3%5D%7B200%7D%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
first solving
![\sqrt[3]{k^{15}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D)
Apply radical rule: ![\sqrt[n]{a^m}=a^{\frac{m}{n}},\:\quad \mathrm{\:assuming\:}a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%2C%5C%3A%5Cquad%20%5Cmathrm%7B%5C%3Aassuming%5C%3A%7Da%5Cge%200)
![\sqrt[3]{k^{15}}=k^{\frac{15}{3}}=k^5](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bk%5E%7B15%7D%7D%3Dk%5E%7B%5Cfrac%7B15%7D%7B3%7D%7D%3Dk%5E5)
then solving
![\sqrt[3]{200}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B200%7D)
prime factorization: 200: 2³ · 5²
![=\sqrt[3]{2^3\cdot \:5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%5Ccdot%20%5C%3A5%5E2%7D)
Apply radical rule:
![=\sqrt[3]{2^3}\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B3%5D%7B2%5E3%7D%5Csqrt%5B3%5D%7B5%5E2%7D)
Apply radical rule:
![\sqrt[n]{a^n}=a,\:\quad \:a\ge 0](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%2C%5C%3A%5Cquad%20%5C%3Aa%5Cge%200)
so
![=2\sqrt[3]{5^2}](https://tex.z-dn.net/?f=%3D2%5Csqrt%5B3%5D%7B5%5E2%7D)
Thus, the main expression becomes
![=2k^5\sqrt[3]{25}](https://tex.z-dn.net/?f=%3D2k%5E5%5Csqrt%5B3%5D%7B25%7D)
Therefore, we conclude that:
Hence, option B is correct.
Answer:
Yes, -13 is less than f/2.
Step-by-step explanation:
Since we are given a fraction with a variable, it may seem hard at first to find out which number is greater.
First, we have to realise that -13 is a negative number. Any negative number is always equal to less than a positive number.
Whatever number we replace our variable f with, whether it be 2, 3, 9, or 0.00000000000001, we can rest assure that it will be a <em>positive</em> number.
Since positive numbers are greater than negative ones, -13 has to be less than f/2.
If the fraction was negative (- f/2), this would be a nearly impossible question to answer.
Answer:
<h2>(90Degree)(0.3Degree/1min) = 300min</h2><h2>since 60min = 1hr</h2><h2>(300min)(1hr/60min) = 5hrs</h2>
