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22ilyas22
@22ilyas22
July 2022
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график функций y=kx+7 проходит через точку A, найдите значение k,если: А(1;9) А(-2;1) А(2;8) А(3;4)?
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jekadva
======A(1;9)
9=k*1+7
9=k+7
k=9-7
k=2
======A(-2;1)
1=k*(-2)+7
1=-2k+7
2k=7-1
2k=6
k=3
=======(2;8)
8=k*2+7
8=2k+7
2k=8-7
2k=1
k=0,5
======A(3;4)
4=k*3+7
4=3k+7
3k=4-7
3k=-3
k=-1
2 votes
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Answers & Comments
9=k*1+7
9=k+7
k=9-7
k=2
======A(-2;1)
1=k*(-2)+7
1=-2k+7
2k=7-1
2k=6
k=3
=======(2;8)
8=k*2+7
8=2k+7
2k=8-7
2k=1
k=0,5
======A(3;4)
4=k*3+7
4=3k+7
3k=4-7
3k=-3
k=-1