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2 years ago

Explain the difference between solving 2x < 7 and solving -2x < 7 .

2 answers:

7 0

You can simply the above equation
1. 2x<7 = x<3.5 which means in order to satisfy the above notation the value of x has to be less that 3.5
2. -2x<7 = x<-3.5 it means in order to satisfy this notation the value of x has to be more than 3.5 in magnitude having - sign ..

zhuklara [117]2 years ago

6 0

When solving inequalities, dividing by a negative number requires you to fliip the sign.

So to solve 2x < 7 you simply divide by 2

x < 7/2

However with -2x < 7, you have to flip the sign

x > -7/2

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BARSIC [14]

Answer:

Step-by-step explanation:

well less than 3

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2 years ago

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On a local sports team 20% of 50 players are left-handed how many left-handed players are on the team

mr Goodwill [35]

Answer:

10 players

Step-by-step explanation:

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2 years ago

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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 $a^{81} b^{91} c^{27}$

Blizzard [7]

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

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

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

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

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

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

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

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2 years ago

You move down 6 units. You end at (10, 4). Where did you start?

Alexandra [31]

Answer:

(10,10)

Step-by-step explanation:

coordinates are (x,y)

x is horizontal(left and right), y is vertical (up and down)

going down 6 means subtracting 6 from the second number

lets reverse that to go up and add 6 back

4+6=10

so the original point was (10,10)

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2 years ago

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nydimaria [60]

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3 years ago