4. A cruise ship travels in the direction of 55degrees for 40 miles, then changes course to a direction of 100 degrees for 35 mi
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Answer:
(C)Determine the principal square root of both sides of the equation.
Step-by-step explanation:
Given: Isosceles right triangle XYZ (45°–45°–90° triangle)
To Prove: In a 45°–45°–90° triangle, the hypotenuse is times the length of each leg.
Proof:
Because triangle XYZ is a right triangle, the side lengths must satisfy the Pythagorean theorem,
Since a=b in an isosceles triangle:
Therefore, the next step is to Determine the principal square root of both sides of the equation.
If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial. C is our constant, 2, and d is the leading coefficient, 1. The factors of 2 are +/- 1 and +/-2. The factors for 1 are +/-1 only. Meaning, in all, there are 4 possibilities as roots for this polynomial. But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above. Let's try x = -1, factor form (x + 1). If there is no remainder when we do the synthetic division, then -1 is a root. Put -1 outside the "box" and the coefficients from the polynomial inside: -1 (1 2 -1 -2). Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1. Put that -1 up under the 2 and add to get 1. Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2. -1 times -2 is 2, and -2 + 2 = 0. So we have our first root of (x+1). The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!). The new polynomial is 
. That can also be factored to find the remaining 2 roots. Use standard factoring to find that the other 2 solutions are (x+2) and (x-1). Our solutions then are x = -2, -1, 1, choice B from above.