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raytiller1
A tower in Centerville sends radio signal a certain distance (in miles) according to the equation: x2 + y2 = 3,600. The Martinez family lives 50 miles north and 70 miles east of Centerville. Part 1: Can the Martinez family receive the radio signal from Centerville?
Do you know the Pythagorean theorem
You're given the equation:\[x^2+y^2=3600\]and you should know this will be a right triangle where \[a^2+b^2=c^2\]Therefore first figure out what c is...that will be the range of the signal: \[3600 = c^2\]\[c=60 \space miles\]The problem states that the Martinez family lives 50 miles north and 70 miles east of Centerville...which means a = 50 and b = 70...what do you get for C using the same equation as above (Pythagorean Theorem)?
ok so 50^2+70^2=60^2 ?@Shane_B 2500+4900= 3600 ?
and so no they can't get the signal?
I think you got confused there. \[50^2+70^2= c^2\]You have to solve for c in that equation to get how far they are from the signal. If it's greater than 60 miles, they won't receive it.
i thought we solved for c already?
I first solved for c just to get the range of the signal based on x^2+y^2=c^2=3600. Then you have solve for how far they are away from it...using the same base equation of \[a^2+b^2=c^2\]
I don't want to confuse you more but since the question is only asking if their distance is greater than the signal's distance, you could essentially boil it down to this question: Is 3600 < 50^2+70^2?
Maybe that will make more sense to you
If the above equation is true, they will get the signal.
yes ok now i get it i think so 50^2 +70^2 > 3600 so they will get it?
If it's > than 3600...they are too far away...it must be <=
so yes they wont get it because 25000 + 49000=74000 >3600
That's correct. Maybe this will help clear this up. The range of the radio signal is:\[\sqrt{3600}=60 \space miles\]The distance the family is from it is:\[\sqrt{50^2+70^2}=86 \space miles\]
You can answer the question using either method...I probably should have started with the simpler one: 50^2+ 70^2 > 3600 so they are out of range.
ok i have another question...
am i right? A store displays six computers on a shelf side–by–side. If the first computer is eight inches wide and each successive computer is four inches wider than the previous one, find the total width of the computers on the shelf. Part 1: Describe the sigma notation used in answering the question above. (2 points) Part 2: Show all your work and answer the question. (3 points) a few moments agoEdit Question 8 ∑ 4 + 4n n=1 ∑ 4 + 4n = 4 ∑ 1 + n 4 * 8 + 4 ∑ n 32 + 4 [8 * 9 / 2] 32 + 40 72 inches