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tekilochka [14]
3 years ago
9

Point C has the coordinates (–1, 4) and point D has the coordinates (2, 0). What is the distance between points C and D?

Mathematics
2 answers:
QveST [7]3 years ago
6 0
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
C&({{ -1}}\quad ,&{{ 4}})\quad 
%  (c,d)
D&({{ 2}}\quad ,&{{ 0}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
CD=\sqrt{[2-(-1)]^2+(0-4)^2}\implies CD=\sqrt{(2+1)^2+(-4)^2}
\\\\\\
CD=\sqrt{9+16}\implies CD=\sqrt{25}\implies CD=5
Tema [17]3 years ago
6 0
Use the distance formula: sqrt[(x1-x2)^2 - (y1-y2)^2]

x1: -1
x2: 2
y1: 4
y2: 0

Answer: 5
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Anon25 [30]

m∠X + m∠Y < 90° is true.

<h3>How to solve it?</h3>

XYZ is a triangle.

We have to find the option that is true about △XYZ.

By angle sum property of a triangle,

The sum of all the interior angles of a triangle is always equal to 180°.

So, ∠X + ∠Y + ∠Z = 180°

Given, m∠Z > m∠X + m∠Y.

This implies that ∠Z is greater than the sum of the angles ∠X and ∠Y.

so, ∠X + ∠Y must be less than 90°.

Hence, we can say that ∠X + ∠Y < 90°

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bagirrra123 [75]

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