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UNO [17]
2 years ago
7

Find the distance between the points 4,7 and -5,7

Mathematics
2 answers:
SashulF [63]2 years ago
6 0

<u>Distance formula is</u>: \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2 }

       (x_{1},y_{1}) = (4,7)\\(x_{2},y_{2}) = (-5,7)

Thus let us plug in all the know values and solve:

      \sqrt{(-5-4)^2+(7-7)^2} =\sqrt{(-9)^2} =\sqrt{81} = 9

Thus the <u>distance is 9</u>

<u></u>

Hope that helps!

GenaCL600 [577]2 years ago
3 0

Answer:

9

Step-by-step explanation:

Horizontal line from 4 to -5.

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A line passes through (10,-1) and has a slope of -1/2 whats the equation of the line
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Answer:

Y=-1/2x+4

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-1=-1/2(10)-b

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4=b

8 0
3 years ago
What is the congruence correspondence, if any, that will prove the given triangles congruent?
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Answer:

<h3>HL</h3>

Step-by-step explanation:

The congruence correspondence between the two given triangles is HL as one leg and hypotenuse are given equal.

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Help 20 pts A string is tied to the top of a plant to keep it stabilized. The plant is 5 feet tall and the string is connected t
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The string tied to the plant would most likely be 7 feet long
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3 years ago
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Maksim231197 [3]

Answer:

3.3

Step-by-step explanation:

We determine the plane formed by u_1 and u_2. The normal to the plane is given by their cross product:

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\begin{pmatrix} - 2\\ - 4\\ 1 \end{pmatrix}\times\begin{pmatrix} - 4\\ 1 \\ - 4\end{pmatrix}=\begin{pmatrix} 15 \\ - 12\\ - 18\end{pmatrix}

The equation of the plane is then given by

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The distance between a vector \begin{pmatrix} x_1 \\ y_1 \\ z_1 \end{pmatrix} and a plane Ax+By+Cz +D =0 is given by

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d=\dfrac{|(15\times3)+(-12\times-8)+(-18\times3)+0|}{\sqrt{15^2+(-12)^2+(-18)^2}}

d=\dfrac{|45+96-54|}{\sqrt{225+144+324}}=\dfrac{87}{\sqrt{693}}

d =\dfrac{87}{26.32\ldots}\approx 3.3

4 0
3 years ago
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