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Anuta_ua [19.1K]
3 years ago
6

Write an equation of the line containing point (5,-4) and parallel to line x=2.

Mathematics
1 answer:
Nikolay [14]3 years ago
3 0
X=5 because it is parallel to x=2 and also include the point (5,-4)
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What is the sum of 2,480 - 170
Natalija [7]

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2,310

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
1 2/5m - 3/5 ( 2/5m + 1 )
Hunter-Best [27]

Answer:

1 4/25 - 3/5, or 29/25 - 3/5, or \frac{(29m - 15)}{25}

it doesn't really matter they all mean the same

Step-by-step explanation:

1 2/5m - 3/5(2/5m + 1)

multiply first

1 2/5m - 6/25m - 3/5

multiply 1 2/5 only so the denominator is 25 as we can't simplify 6/25, it's so we can add them together, we can't add 3/5 tho bc it's a different term so there's no need to multiply that.

1 10/25m - 6/25m - 3/5

add/subtract

1 4/25m - 3/5

4 0
3 years ago
Help! Exponential Equation WITHOUT CALCULATOR
ale4655 [162]

1.
(2^{x} -2^{-x})^{2} = 4^{2} \\ \\2^{2x}-2*2^{x}*2^{-x} + 2^{-2x} =4^{2} \\ \\2^{2x}-2*2^{x-x} + 2^{-2x} =4^{2} \\ \\ 2^{2x}+ 2^{-2x} - 2=16 \\ \\ 2^{2x}+ 2^{-2x} = 18

\\ \\ 2.( 2^{x})^{3} - (2^{-x}^) ^{3}= (2^{x} -2^{-x})(2^{2x}+2^{x}*2^{-x}+2^{-2x})
\\ \\ ( 2^{x})^{3} - (2^{-x}^)^{3}=(2^{x} -2^{-x})(2^{2x}+2^{-2x}+2^{x}*2^{-x})
\\ \\ ( 2^{x})^{3} - (2^{-x}^)^{3}=(2^{x} -2^{-x})(2^{2x}+2^{-2x}+1)
\\ \\ ( 2^{x})^{3} - (2^{-x}^)^{3}=(2^{x} -2^{-x})((2^{2x}+2^{-2x})+1)


\\ \\ ( 2^{x})^{3} - (2^{-x})^{3} =(4)((18)+1)= 76
\\ \\ 2^{3x}-2^{-3x} =76

4 0
3 years ago
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