![\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) ~\hfill a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \boxed{a^6+b^6}\implies a^{2\cdot 3}+b^{2\cdot 3}\implies (a^2)^3+(b^2)^3 \\[2em] [a^2+b^2] [(a^2)^2-a^2b^2+(b^2)^2]\implies \boxed{(a^2+b^2)(a^4-a^2b^2+b^4)}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdifference%20and%20sum%20of%20cubes%7D%20%5C%5C%5C%5C%20a%5E3%2Bb%5E3%20%3D%20%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29%20~%5Chfill%20a%5E3-b%5E3%20%3D%20%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cboxed%7Ba%5E6%2Bb%5E6%7D%5Cimplies%20a%5E%7B2%5Ccdot%203%7D%2Bb%5E%7B2%5Ccdot%203%7D%5Cimplies%20%28a%5E2%29%5E3%2B%28b%5E2%29%5E3%20%5C%5C%5B2em%5D%20%5Ba%5E2%2Bb%5E2%5D%20%5B%28a%5E2%29%5E2-a%5E2b%5E2%2B%28b%5E2%29%5E2%5D%5Cimplies%20%5Cboxed%7B%28a%5E2%2Bb%5E2%29%28a%5E4-a%5E2b%5E2%2Bb%5E4%29%7D)
about the second one... well, is a "fait accompli" that using the pythagorean theorem, if x = 8 and y = 5, the hypotenuse must be √(8² + 5²) = √(89), which is neither of those choices.
5, 8, 13 are no dice, namely 5² + 8² ≠ 13
25, 64, 17 is are no dice too, because 25² + 17² ≠ 64²
however, 5,12 and 13 are indeed a pythagorean triple
also is 39, 80, 89.
when looking for a pythagorean triple, recall that c² = a² + b².
so the longest leg is the sum of the square of the small ones.
so what you'd do is, check the small legs, square them, add them up, if they're indeed a pythagorean triple, they "must" add up to the longest leg.
Answer: 11
Step-by-step explanation: You divide 495/45 and you get 11
Answer: 10
Step-by-step explanation:
You times 4 and 5 then you divide in half because it is a triangle
Answer:
d=1.4
Step-by-step explanation:
cuz 10-1.4=8.6
Answer:
$350,000
Step-by-step explanation:
Let's define:
- s: amount of short-range missiles produced
- m: amount of medium-range missiles produced
- l: amount of long-range missiles produced
From the total production and the ratios we can write the following equations:
s + m + l = 3000
s/m = 3/3 = 1 = m/s
s/l = 3/4 or l/s = 4/3
Dividing the first equation by s, we get:
s/s + m/s + l/s = 3000/s
1 + 1 + 4/3 = 3000/s
10/3 = 3000/s
s = 3000*3/10 = 900
m = 900
l = 4/3*900 = 1200
From the money that the countries plans to use and each missile cost, we can write the following equation:
200,000*s + 300,000*m + cost*l = 870,000,000
Replacing with previous result:
200,000*900 + 300,000*900 + cost*1200 = 870,000,000
cost = (870,000,000 - 200,000*900 - 300,000*900)/1200 = 350,000
