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A square is cut into three equal rectangles. How does the area of the square compare to the sum of the areas of the three rectan
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This is an interesting question. I chose to tackle it using the Law of Cosines.
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
AC² = AB² + BC² - 2·AB·BC·cos(B)
AM² = AB² + MB² - 2·AB·MB·cos(B)
Subtracting twice the second equation from the first, we have
AC² - 2·AM² = -AB² + BC² - 2·MB²
We know that MB = BC/2. When we substitute the given information, we have
8² - 2·3² = -4² + BC² - BC²/2
124 = BC² . . . . . . . . . . . . . . . . . . add 16, multiply by 2
2√31 = BC ≈ 11.1355
Answer:
The quotient is
Step-by-step explanation:
The given polynomial division is
To perform the synthetic division we write out the coefficients and arrange them as shown.
1 -4 4 -1
<u>1 | 1 -3 1</u>
1 -3 1 0
The quotient is
The descending powers of x means from highest degree to the least
The remainder is 0