9 x 2 = 18
6 x 3 = 18
and 18 x 1 = 18
<span>the statement that contain simile is : D. i think that i shall never see a poem lovely as a tree
Simile is a writing technique which compares a thing with another thing from a different kind.
In sentence D, the writer compare a 'poem' which is a form of artwork, with a 'tree' which is a form of living organism</span>
Add the two and you should get 2 and 7/16
The question is incomplete. Here is the complete question:
Mr.yueng graded his students math quizzes students came up with four different answers when solving the equation x3=22. Which answers is correct.
(A) 
(B) ![\sqrt[3]{22}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B22%7D%20)
(C) ![\pm \sqrt[3]{22}](https://tex.z-dn.net/?f=%5Cpm%20%5Csqrt%5B3%5D%7B22%7D%20)
(D) 
Answer:
(B) ![\sqrt[3]{22}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B22%7D)
Step-by-step explanation:
Given:
The equation to solve is given as:
Here, the left hand side of the equation has a variable 'x' in exponent form. So, in order to solve for 'x', we have to eliminate the exponent.
For removing the exponent, we have to take cubic root on both the sides. As we know that,
![\sqrt[n]{x^n} =x](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5En%7D%20%3Dx)
So, taking cubic root on both the sides, we get
![\sqrt[3]{x^3}=\sqrt[3]{22}\\\\x=\sqrt[3]{22}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E3%7D%3D%5Csqrt%5B3%5D%7B22%7D%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B22%7D)
Therefore, the second student has written the correct answer and hence the correct option is (B).
By using the rules that the value inside square root can’t be negative and the denominator value can’t be zero, the domain for the given function is a) x<-1 and x>1 b) p≤1/2 c) s>-1.
I found the complete question on Chegg, here is the full question:
Write the restrictions that should be imposed on the variable for each of the following function. Then find, explicitly, the domain for each function and write it in the interval notation a) f(x)=(x-2)/(x-1) b) g(p)=√(1-2p) c) m(s)= (s^2+4s+4)/√(s+1)
Ans. We know that a number is not divisible by zero and number inside a square root can not be negative. In both the cases the outcome will be imaginary.
a) For this case the denominator x-1 can not be zero. So, x ≠1 and the domain is x<-1 and x>1.
b) For this case the value inside square root can’t be negative. So, p can’t be greater than 1/2 the domain is p≤1/2.
c) For this case also the value inside square root can’t be negative and the denominator value can’t be zero. So, s can’t equal or less than -1 and domain is s>-1.
Learn more about square root here:
brainly.com/question/3120622
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