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Natasha2012 [34]
3 years ago
13

Helpppppppppppppppppppppppppppppppppppp

Mathematics
1 answer:
yanalaym [24]3 years ago
6 0

Answer:

b

Step-by-step explanation:

a^3b^2c = b(a^3bc)

5bc = b(5c)

15a^2b = b(15a^2)

GCF = b

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What is the area of a sector with a central angle 210 and a diameter of 4.6
gizmo_the_mogwai [7]
Theta = 210°
r = 4.6/2 = 2.3
Area of circle = pi * r * r
= 3.14 * 2.3 * 2.3 = 16.61
Area of sector =
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What is the coefficient of the term 4/3yz ?
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The coefficient is 4/3
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In a linear Pair angles are?
tatiyna

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):
zlopas [31]
<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)

First pair of opposite sides:

<u>Slope:</u>

\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}

Second pair of opposite sides:

<u>Slope:</u>

\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\

So the diagonals measure the same, therefore this is a rectangle.

5 0
3 years ago
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