Given the a^b=x, evaluate the following. a^b+2
1 answer:
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Answer:
Step-by-step explanation:
1²+16²=1+256=257≠289(17²)
2²+16²=4+256=260≠17²
3²+16²=9+256=265≠17²
4²+16²=16+256=272≠17²
5²+16²=25+256=281≠17²
6²+16²=36+256=292>17²
6²+15²=36+225=261≠17²
7²+15²=49+225=274≠17²
8²+15²=64+225=289=17²
9²+14²=81+196=277≠17²
10²+14^2=100+196>289
10²+13²=100+169=269≠17²
11²+13³=121+169=290>17²
11²+12²=121+144=265≠17²
12²+12²=144+144=288≠17²
so only integer values are 8²+15²=17²
(x+4)²+(y-7)²=17²
if x+4=8
x=8-4=4
y-7=15
y=15+7=22
so q is (4,22)
if x+4=15
x=15-4=11
y-7=8
y=8+7=15
point is (11,15)
