Can anyone answer this please i need to be for sure I’m right

Mathematics

1 answer:

Genrish500 [490]3 years ago

7 0

The answer to this question is B, (-infinity, -28].

We can get this answer by first multiplying each side of the inequality by 7. That would get rid of the fraction. When one does that, the result is d + 28 $\leq$ 0. That means that d $\leq$ -28. In interval notation, which is the notation the problem is asking us, that would be (-infinity, -28], since d is all values less than -28 this includes infinity, but it also includes -28, so there is a ] around it. That means that the answer to this question is B, (-infinity, -28].

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Given $a \equiv 1 \pmod{7}$, $b \equiv 2 \pmod{7}$, and $c \equiv 6 \pmod{7}$, what is the remainder when $a^{81} b^{91} c^{27}$

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$\begin{cases}a\equiv1\pmod7\\b\equiv2\pmod7\\c\equiv6\pmod7\end{cases}$

$a^{81}\equiv1^{81}\equiv1\pmod7$

$b^{91}\equiv2^{91}\pmod7$

$2^{91}\equiv(2^3)^{30}\times2^1\equiv8^{30}\times2\pmod7$
$8\equiv1\pmod7$
$2\equiv2\pmod7$
$\implies2^{91}\equiv1^{30}\times2\equiv2\pmod7$

$c^{27}\equiv6^{27}\pmod7$

$6^{27}\equiv(-1)^{27}\equiv-1\equiv6\pmod7$

$\implies a^{81}b^{91}c^{27}\equiv1\times2\times6\equiv12\equiv5\pmod7$