1. ## algebra problems

1. Pic 11. What are the possible values for x?

A. x > 3.75
B. x < 2.5
C. 0 < x < 2.5
D. 2.5< x < 3.75

2. Pic 16. Two boats start at the dock. The 1st sails 4 miles due west, then turns 46 degrees south and sails 1.8 miles. The 2nd sails 4 miles due east, then turns 58 degrees north and sails 1.8 miles. Which boat is farther from the dock?

A. the 1st
B. the 2nd

2. Originally Posted by dgenerationx2
1. Pic 11. What are the possible values for x?

A. x > 3.75
B. x < 2.5
C. 0 < x < 2.5
D. 2.5< x < 3.75
The sum of any two sides of a triangle must be greater than the third side.

I tried all three combinations just in case, but your drawing seems to be accurate. The sum of the two shorter sides must be greater than the longest side.

$(4x)+(6x-20)>(2x+10)$

$10x-20>2x+10$

$8x>30$

$x>3.75$

Originally Posted by dgenerationx2
2. Pic 16. Two boats start at the dock. The 1st sails 4 miles due west, then turns 46 degrees south and sails 1.8 miles. The 2nd sails 4 miles due east, then turns 58 degrees north and sails 1.8 miles. Which boat is farther from the dock?

A. the 1st
B. the 2nd
The first boat travels 4 mi on one leg of the triangle and 1.8 miles on another. The angle between these legs is 180 - 46 = 134 degrees.

The second boat travels 4 mi on one leg of its triangle and 1.8 miles on another. The angle between these legs is 180 - 58 = 122 degrees.

Since the two triangles have angles flanked by legs of equal length, we need only use the largest angle to determine the longest 3rd side. The largest angle was produced by the first boat, so this one is furthest from the dock.

3. Hello, dgenerationx2!

2. Two boats start at the dock.
The 1st sails 4 miles due west, then turns 46° south and sails 1.8 miles.
The 2nd sails 4 miles due east, then turns 58° north and sails 1.8 miles.
Which boat is farther from the dock?
Code:
                                    * B
* /
*   /1.8
*     /
P    4       *   122°/58°
W - - - * - - - - o - - - - * - - - E
46°/134°   *  D    4    Q
/     *
1.8 /   *
/ *
A *

The 1st boat starts at $D$, sails 4 miles west to $P$,
. . then turns 46° south and sails 1.8 miles to $A.$
$\angle WPA = 46^o \quad\Rightarrow\quad \angle DPA = 134^o$

The 2nd boat starts at D, sails 4 miles east to Q,
. . then turns 58° north and sails 1.8 miles to $B.$
$\angle BQE = 58^o \quad\Rightarrow\quad \angle DQB = 122^o$

We will use the Law of Cosines . . .

In $\Delta APD\!:\;\;AD^2\:=\:1.8^2 + 4^2 - 2(1.8)(4)\cos134^o \:=\:29.24308053$
. . Hence: . $D \:\approx\:5.4$ miles.

In $\Delta DQB\!:\;\;DB^2 \:=\:4^2 + 1.8^2 - 2(4)(1.8)\cos122^o \:=\:26.8708324$
. . Hence: . $DB \:\approx\:5.2$ miles.

Therefore, the first boat is farther from the dock.