Question:
What are the solution(s) to the quadratic equation 50 – x2 = 0?
A) x = ±2Plus or minus 2 StartRoot 5 EndRoot
B) x = ±6Plus or minus 6 StartRoot 3 EndRoot
C) x = ±5Plus or minus 5 StartRoot 2 EndRoot
D) no real solution
Answer:
C) x = ±5Plus or minus 5 StartRoot 2 EndRoot
Isolating 2abCos(c) on one side of the equation and using the given values of a, b and c we can find the answer to this question as shown below:
<span>The answers are b = 5 square root of 3; b = -5 square root of 3. f(b) = b^2 – 75. If f(b) = 0, then b^2 – 75 ) 0. b^2 = 75. b = √75. b = √(25 * 3). b = √25 * √3. b = √(5^2) * √3. Since √x is either -x or x, then √25 = √(5^2) is either -5 or 5. Therefore. b = -5√3 or b = 5√3.</span>
Answer:
see below (I hope this helps!)
Step-by-step explanation:
First of all, we can clearly tell that a straightedge was used to create CD but that doesn't prove anything so we can eliminate the second and fourth options. Upon measuring the lengths of AC and CB, we see that they are equal so the answer is Yes; the compass was kept at the same width to create the arcs for points C and D. The reason why this is so important is because we know that AC = CB = BD = DA because the compass width is the same, so therefore, ACBD is a rhombus based on the definition of a rhombus. We know that the diagonals of a rhombus bisect each other and since CD and AB are diagonals, we know that CD bisects AB.
