One half a number increased by three is greater than zero or less than or equal to negative three
2 answers:
Answer:
The required inequality for both the statement is: and
The solution for the inequalities are x>-6 and respectively.
Step-by-step explanation:
Consider the provided information.
One half a number increased by three is greater than zero
Let the number is x.
Thus, the required inequality is:
Or One half a number increased by three is less than or equal to negative three.
Thus, the required inequality is:
Now solve the above inequality for x.
or
or
or
or
Hence, the value of x is or
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So hmm check the picture below
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\qquad \begin{cases} h=height\\ r=radius\\ ------\\ h=2(2r) \end{cases}\implies V=\cfrac{\pi r^2[2(2r)]}{3} \\\\\\ V=\cfrac{4\pi r^3}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ah%3Dheight%5C%5C%0Ar%3Dradius%5C%5C%0A------%5C%5C%0Ah%3D2%282r%29%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20r%5E2%5B2%282r%29%5D%7D%7B3%7D%0A%5C%5C%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B4%5Cpi%20r%5E3%7D%7B3%7D)
the radius "r" is half the diameter, meaning, the diameter is 2r long
now, if the height "h" is twice "d" or 2d, then that means h = 2(2r)
thus