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Maru [420]
3 years ago
10

Filiate Consider the line MU for M(-1, 1) and U(4, 5). what’s the distance from M to U ?

Mathematics
2 answers:
sergey [27]3 years ago
8 0

Answer:

The answer is

<h3>\sqrt{41}  \:  \: or \:  \: 6.403 \:  \:  \: units</h3>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3>d =  \sqrt{ ({x1 - x2})^{2}  +  ({y1 - y2})^{2} }  \\</h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

M(-1, 1) and U(4, 5)

The distance from M to U is

<h3>|MU|   = \sqrt{ ({ -  1 - 4})^{2}  + ( {1 - 5})^{2} }  \\  =  \sqrt{ ({ - 5})^{2}  + ( { - 4})^{2} }  \\  =  \sqrt{25 + 16}</h3>

We have the final answer as

<h3>\sqrt{41}  \:  \: or \:  \: 6.403 \:  \:  \: units</h3>

Hope this helps you

professor190 [17]3 years ago
6 0

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{ \sqrt{41} \:  \:  units}}}}}

Step-by-step explanation:

Let M ( -1 , 1 ) be ( x₁ , y₁ ) and U ( 4 , 5 ) be ( x₂ , y₂ )

<u>Finding</u><u> </u><u>the</u><u> </u><u>distance</u><u> </u><u>from </u><u>M</u><u> </u><u>to</u><u> </u><u>U</u><u> </u>

\boxed{ \sf{distance =  \sqrt{ {(x2 - x1)}^{2} +  {(y2 - y1)}^{2}  } }}

\longrightarrow{ \sf{ \sqrt{ {(4 - ( - 1))}^{2} +  {(5 - 1)}^{2}  } }}

\longrightarrow{ \sf{ \sqrt{ {(4 + 1)}^{2} +  {(5 - 1)}^{2}  } }}

\longrightarrow{ \sf{ \sqrt{ {(5)}^{2}  +  {(4)}^{2} } }}

\longrightarrow{ \sf{ \sqrt{25 + 16}}}

\longrightarrow{ \sf{ \sqrt{41} }} units

The distance from M to U is \sf{ \sqrt{41} \:  \:  units}

Hope I helped!

Best regards! :D

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faltersainse [42]

Answer:

scatter plot A: -0.90

scatter plot B: 0.89

scatter plot C: -0.76

scatter plot D: 0.55

Step-by-step explanation:

if the line is going down, it's negative and if it goes up it's positive. The closer the points are <em>the higher the number</em>.

HOPE THIS HELPS!!

3 0
3 years ago
Can someone please help with this question pleaseee
Ulleksa [173]

Answer: Choice C) 10.5

The distance from A to C is 7 units (count out the spaces between the two points, or subtract y coordinates 4-(-3) = 4+3 = 7)

Let AC = 7 be the base of the triangle. You might want to rotate the image so that AC is laying horizontally rather than being vertical.

Now move to point P. Walk 3 spaces to the right until you land on segment AC. This shows that the height of the triangle is 3 when the base is AC = 7.

base = 7, height = 3

area of triangle = (1/2)*base*height

area of triangle = 0.5*7*3

area of triangle = 10.5 square units

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3 years ago
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<em>-</em><em> </em><em>BRAINLIEST </em><em>answerer</em><em> ❤️</em>

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