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2 years ago

Math Help Part 3 (Please Help)

Mathematics

2 answers:

Bezzdna [24]2 years ago

5 0

Answer:

(a) median

(b) angle bisector

Step-by-step explanation:

The perpendicular bisector of a side of a triangle is a line perpendicular to the side and passing through its midpoint.
The angle bisector of an angle of a triangle is a straight line that divides the angle into two congruent angles.
A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side of the vertex.
An altitude of a triangle is the perpendicular segment from a vertex of a triangle to the opposite side (or the line containing the opposite side).

(a) median

IG = IH ⇒ I is the midpoint GH, BUT it is not perpendicular to FI, so median

(b) angle bisector

m∠XYV = m∠ZYV, BUT YV is not perpendicular to XZ, so angle bisector

m∠BDC = 90° but AD ≠ DC (so not perpendicular bisector)

yulyashka [42]2 years ago

4 0

I wish I could help but I suck at math

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3 years ago

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I have 14 but need the answer to 18.

nignag [31]

Answer:

18.85 mm.

7.73 mm^2.

Step-by-step explanation:

The side of the square = sqrt 36 = 6 mm.

Now this side = the diameter of the circle.

So the circumference of the circle = π * 6 = 18.85 mm.

The answer to the second part

= area of the square - area of the circle

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= 36 - 28.27

= 7.73 mm^2.

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3 years ago

) An engineer deploys a set of sensors. Each sensor reports its location as a point in two-dimensional space. The points are (2,

r-ruslan [8.4K]

Answer:

The polygon area is $51.5 U^{2}$

Step-by-step explanation:

Points:

(2,7), (8,2), (6,11), (6,5), (11,6)

1. The area of the polygon could be calculate with the eqation:

$A=\frac{1}{2}\left[\begin{array}{ccc}x_{1} &y_{1}\\x_{2} &y_{2}\\x_{3} &y_{3}\\x_{4} &y_{4}\\x_{5} &y_{5}\end{array}\right]$

2. Replace the coordinates in equation for A:

$A=\frac{1}{2}\left[\begin{array}{ccc}2&7}\\8&2\\6&11\\6&5\\11&6\end{array}\right]$

3. Solve the determinant, and calculate A:

$A=\frac{1}{2}\left[\begin{array}{ccc}2&7}\\8&2\\6&11\\6&5\\11&6\end{array}\right]\\A=\frac{1}{2}[(6)(7)+(11)(11)+(8)(6)+(6)(2)]-[(2)(11)+(6)(6)+(11)(2)+(8)(5)]\\A=\frac{1}{2}[42+121+48+12]-[22+36+22+40]\\A=\frac{1}{2}[223]-[120]\\\\A=\frac{1}{2}[103]=51,5 U^{2}$