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Nikolay [14]
3 years ago
15

Solve for the distance between the points (4, 4) and (1, -3).

Mathematics
1 answer:
antoniya [11.8K]3 years ago
7 0
I think the distance is 3/7. This is because 4 minus 1 is 3 (they are both in the same quadrant, so u subtract/are both positive too). -3 is three away from zero and 4 is 4 away from zero. If you add those together, you get those points (also, u add numbers in different quadrants together). Therefore, the distance is 3/7. Hope this helps!
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Step-by-step explanation:

The y - intercept is 1 . Since the line goes through the point ( 0 , 2 ) and ( -1 , -2 ) the slope is m = ( - 2 - 2 ) / ( - 1 - 0 ) = -4/-1. so the equation will be y = -4/1x + 1

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3 0
3 years ago
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Please solve this for me using substitution
Vinvika [58]

<h2> <u>Explanation</u><u>:</u></h2>

\bf \underline{★Given-} \\

\textsf{4x + 5y = 8};

\textsf{5x + 4y = 12}

\bf \underline{★To\: find-} \\

\textsf{the value of x and y in equation?}

\bf \underline{★Solution-} \\

\sf \leadsto 4x + 5y = 8 - - - (i)

\sf \leadsto 5x + 4y = 12 - - - (ii)

By first equation,

\sf \leadsto 4x + 5y = 8

\sf \leadsto 4x = 8 - 5y

\sf \leadsto x = \dfrac{8 - 5y}{4}

\textsf{Now, we can find the original value of Y.}\\

\sf \leadsto 5x + 4y = 12

\sf \leadsto 5 \bigg( \dfrac{8 - 5y}{4} \bigg) + 4y = 12

\sf \leadsto \dfrac{40 - 25y}{4} + 4y = 12

\sf \leadsto \dfrac{40 - 25y + 16y}{4} = 12

\sf \leadsto \dfrac{40 - 9y}{4} = 12

\sf \leadsto 40 - 9y = 12(4)

\sf \leadsto 40 - 9y = 48

\sf \leadsto -9y = 48 - 40

\sf \leadsto -9y =  8

\sf \leadsto y = \dfrac{ -8}{9}

\textsf{Now, we can find the original value of X.}\\

\sf \leadsto 4x + 5y = 8

\sf \leadsto  4x + 5 \bigg( \dfrac{ -8}{9} \bigg) = 8

\sf \leadsto 4x - \dfrac{40}{9} = 8

\sf \leadsto \dfrac{36x - 40}{9} = 8

\sf \leadsto 36x - 40 = 8(9)

\sf \leadsto 36x - 40 = 72

\sf \leadsto 36x = 72 + 40

\sf \leadsto 36x = 112

\sf \leadsto x = \dfrac{112}{36}

\underline{\textsf{Answer-}}\\

Therefore, the values of x and y are -8/9 and 112/36 respectively.

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