1. Determine if the lengths 9 cm, 13 cm, and 23 cm can form a triangle.

A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the

Rina8888 [55]

Answer:

Wind speed: $\rm 40\; km \cdot h^{-1}$ .
Speed of the plane in still air: $\rm 320\; km \cdot h^{-1}$ .

Step-by-step explanation:

This problem involves two unknowns:

wind speed, and
speed of the plane in still air.

Let the speed of the wind be $x \rm \; km \cdot h^{-1}$ , and the speed of the plane in still air be $y\rm \; km \cdot h^{-1}$ . It takes at least two equations to find the exact solutions to a system of two variables.

Information in this question gives two equations:

It takes the plane three hours to travel $\rm 960\; km$ from Ottawa to with a tail wind (that is: at a ground speed of $x + y$ .)
It takes the plane four hours to travel $\rm 960\; km$ from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of $-x + y$ .)

Create a two-by-two system out of these two equations:

$\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right.$ .

There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.

Note that this system is equivalent to

$\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

$\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right.$ .

Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only $1 \times 960$ on the right-hand side. In comparison, adding them will give $7 \times 960$ , which is much more complex to evaluate. Subtracting the second equation ( $3 \times (2)$ ) from the first ( $4 \times (1)$ ) will give the equation