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3 years ago

Are the lines y=4x+4 and y=-4x-2 parallel

Mathematics

2 answers:

skelet666 [1.2K]3 years ago

7 0

Answer:

No these are not parallel

Step-by-step explanation:

Parallel lines have the same slope. 4 and -4 are not the same slope, which means they will intersect. A tool to check your answers such as this is Desmos. It really helps. Search Desmos graphing calculator. Hope this helps.

Alex Ar [27]3 years ago

6 0

Answer:

No they are not, they intersect at some point.

Step-by-step explanation:

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There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac

Nataliya [291]

You have a single sequence of integers $a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7$ such that

$\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},$

where $0 \le a_i < i$ for $i = 2, 3, \dots, 7.$

1. Multiply by 7! to get

$\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.$

By Wilson's theorem,

$a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.$

2. Then write $a_7$ to the left and divide through by 7 to obtain

$6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.$

Repeat this procedure by $\mod 6:$

$a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.$

And so on:

$5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.$

Answer: $a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.$

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3 years ago