Please help! I'm having trouble solving this.

Mathematics

1 answer:

Vanyuwa [196]4 years ago

3 0

1st step: break each of these arrows (vectors) into their x- and y-components. The pedestrian vector would be <6,0>.

The wind vector would be <-13cos(45), 13sin(45)> (Notice the x-component is negative because it is to the left.)

2nd step. Add the two vectors: <6, 0> + <-9.1924,9.1924> = <-3.1924, 9.1924> This is the resultant vector in rectangular coordinates. Now you need the magnitude of this vector.

3rd step: Magnitude is always positive and it is sqrt[x-coord2 + y-coord2]

Magnitude = sqrt(94.6915598949) = 9.7309588374 rounded to whatever

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The coordinates of quadrilateral PQRS are P(–3, 0), Q(0, 4), R(4, 1), and S(1, –3). What best describes the quadrilateral?

denis-greek [22]

Answer:

Square

Step-by-step explanation:

Plot the vertices of the quadrilateral PQRS on the coordinate plane (see attached diagram). The diagram shows that this is a square. Let's prove it.

1. Find all sides lengths:

$PQ=\sqrt{(-3-0)^2+(0-4)^2}=\sqrt{(-3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\QR=\sqrt{(0-4)^2+(4-1)^2}=\sqrt{(-4)^2+(3)^2}=\sqrt{16+9}=\sqrt{25}=5\\ \\RS=\sqrt{(4-1)^2+(1-(-3))^2}=\sqrt{(3)^2+(4)^2}=\sqrt{9+16}=\sqrt{25}=5\\ \\SP=\sqrt{(1-(-3))^2+(-3-0)^2}=\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5$

All sides have the same lengths.

2. Find the slopes of all lines:

$PQ:\ \dfrac{4-0}{0-(-3)}=\dfrac{4}{3}\\ \\QR:\ \dfrac{1-4}{4-0}=-\dfrac{3}{4}\\ \\RS:\ \dfrac{-3-1}{1-4}=\dfrac{4}{3}\\ \\SP:\ \dfrac{0-(-3)}{-3-1}=-\dfrac{3}{4}$