Answer:
-1
Step-by-step explanation:
f(0) = -5^0 = -1
Since we know that he ate 1/4 pound one time, and 1/6 another time, we have to add up the fractions.
1/4+1/6
LCM of 4 and 6 = 12
3/12 + 2/12
Add the numerators:
5/12
So, he ate 5/12 of a pound of bars in both breaks.
Hope this helps!
You can solve this one by adding their rates, but you have to invert the numbers they give you to find the rate. Rate is measured in something per unit of time, and "per" indicates division. So the rates are: 1/3 of an hour (Sally) and 1/6 of an hour (Steve). So add the rates: 1/3 + 1/6 = 1/2. (1/3 is 2/6, and 2/6 + 1/6 = 3/6). Since their combined rate is 1/2 of a room in one hour, it takes them two hours to paint the room.
This answer makes sense: It takes Sally three hours by herself; with Steve's help, she's faster, but not THAT much faster, because he's pretty slow. You can often spot-check word problem answers like this by giving it a common sense once-over. If we came up with four hours, that would clearly be wrong: it should take her LONGER if she's getting help.
"determine the location" or namely, is it inside the circle, outside the circle, or right ON the circle?
well, we know the center is at (1,-5) and it has a radius of 5, so the distance from the center to any point on the circle will just be 5, now if (4,-1) is less than that away, is inside, if more than that is outiside and if it's exactly 5 is right ON the circle.
well, we can check by simply getting the distance from the center to the point (4,-1).
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{center}{(\stackrel{x_1}{1}~,~\stackrel{y_1}{-5})}\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d = \sqrt{[4-1]^2+[-1-(-5)]^2}\implies d=\sqrt{(4-1)^2+(-1+5)^2} \\\\\\ d = \sqrt{3^2+4^2}\implies d =\sqrt{9+16}\implies d=\sqrt{25}\implies \stackrel{\textit{right on the circle}}{d = 5}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%5Cstackrel%7Bcenter%7D%7B%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B-5%7D%29%7D%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B-1%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B%5B4-1%5D%5E2%2B%5B-1-%28-5%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%284-1%29%5E2%2B%28-1%2B5%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d%20%3D%20%5Csqrt%7B3%5E2%2B4%5E2%7D%5Cimplies%20d%20%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20d%3D%5Csqrt%7B25%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bright%20on%20the%20circle%7D%7D%7Bd%20%3D%205%7D)
Answer
Algebra
Step-by-step explanation:
Intermediate is the same as algebra 2
