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raketka [301]
3 years ago
8

Pls help soon!Find the distance between the points (-4, -5) and (3, -1).

Mathematics
2 answers:
Harrizon [31]3 years ago
7 0

Answer:

\sqrt{65}

Step-by-step explanation:

Please see work attached.

hit the heart button :)

vaieri [72.5K]3 years ago
5 0

Answer:  \sqrt{65}

Step-by-step explanation:

distance formula is \sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1} )^{2}}

take the 2points in the given (-4,-5) as ((x_{1},y_{1}  )) and (3,-1) as (x_{2},y_{2})

and just substitute to get \sqrt{65}

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Check each equation whose graph is the line that
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Answer:

Points P ( 4 , - 7 ) and  Q ( 1 , 5 ) belong to the equations:

1  ;  4  ;  and  5

Step-by-step explanation:

Equations 1 ; 4 and 5 are the same equation

Equation 1    y  =  - 4*x + 9

Equation 4

y + 7 = - 4 * ( x - 4 )   ⇒  y + 7  = - 4*x + 16   ⇒ y = - 4*x - 7 + 16

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Equation 5

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Now for the equation y = - 4*x + 9

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For x = 4    y = - 4*(4) + 9     ⇒  y = - 16 +  9    ⇒  y = - 7

Then point P is in the line  y = - 4*x + 9

Point Q (1 , 5 )

For x = 1    y = - 4 * ( 1) + 9     ⇒  y = - 4 + 9    ⇒  y = 5

Point Q is in the line y = - 4*x + 9

Equation 2

y = - 4*x - 23

Point P ( 4 , - 7 )

For x = 4       y  = 16 - 23   y  = - 7

Point P is in the line

Point Q

For x = 1     y = - 4 *(1) - 23      ⇒   y = - 27

Then poin Q is not in the line

Equation 3

y - 1 = - 4 * ( x - 4 )

y  - 1 =  -4*x + 16    ⇒   y  = - 4*x + 17

Point P ( 4 , - 7 )

For x = 4

y = - 16 + 17   ⇒ y = 1    

Point P is not in the line

And Point Q ( 1 , 5 )

For  x = 1

y = - 4* ( 1 ) + 17    ⇒  y  = 13    Q is not in the line

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