Answer:
Emma was right to say that both packages are close to 2 pounds.
If we round-off the weights of both to the nearest whole number we would get 2 pounds on both. We look at the the difference of both from 2 pounds we get the following:
2.36 - 2.0 = 0.36
2.09 - 2.0 = 0.09
The smaller the difference the closer it is to the base value. So we can say that the closest one is 2.0 pounds is the package that weighs 2.09 pounds.
Answer:
A. <em>A </em>and <em>F</em>.
Step-by-step explanation:
By observing the number line given, you can assume that in between -4 and -6 would be the number -5, replaced by the variable <em>A</em>, and the same thing with 4 and 6, where 5 is replaced by <em>F</em>.
Answer:
what do you want to ask brother...
Step-by-step explanation:
or you post this by mistake.....
well, if the diameter is 5, thus its radius must be half that, or 2.5, and therefore, the radius of the one four times as much will be (4)(2.5).
Let's simply get their difference, since that'd be how much more is needed from the smaller to larger sphere.
![~\hfill \stackrel{\textit{surface area of a sphere}}{SA=4\pi r^2}\qquad \qquad r=radius~\hfill \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large difference of their areas}}{\stackrel{\textit{radius of (4)(2.5)}}{4\pi (4)(2.5)^2}~~ - ~~\stackrel{\textit{radius of 2.5}}{4\pi (2.5)^2}}\implies 100\pi -25\pi \implies 75\pi ~~ \approx ~~235.62~ft^2](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7Bsurface%20area%20of%20a%20sphere%7D%7D%7BSA%3D4%5Cpi%20r%5E2%7D%5Cqquad%20%5Cqquad%20r%3Dradius~%5Chfill%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20difference%20of%20their%20areas%7D%7D%7B%5Cstackrel%7B%5Ctextit%7Bradius%20of%20%284%29%282.5%29%7D%7D%7B4%5Cpi%20%284%29%282.5%29%5E2%7D~~%20-%20~~%5Cstackrel%7B%5Ctextit%7Bradius%20of%202.5%7D%7D%7B4%5Cpi%20%282.5%29%5E2%7D%7D%5Cimplies%20100%5Cpi%20-25%5Cpi%20%5Cimplies%2075%5Cpi%20~~%20%5Capprox%20~~235.62~ft%5E2)
To add a mixed number you must first convert it into a heavy headed fraction eg. 12/4
Then you must find a common denominator for all of the fractions like you normally do for regular addition of fractions
After you have added them you must convert the heavy headed fraction back into a mixed number by seeing how many times the numerator goes into the denominator
