The quadratic equation (3k-2)x^2 +12x+3(k+1) =0 has equal roots. Find the two possible values of k.

Answers

Answer 1

Answer: $x_1=x_2 \Rightarrow \Delta=0\n\Delta=12^2-4\cdot(3k-2)\cdot3(k+1)\n\Delta=144-(36k-24)(k+1)\n\Delta=144-36k^2-36k+24k+24\n\Delta=-36k^2-12k+168\n-36k^2-12k+168=0\n-3k^2-k+14=0\n-3k^2+6k-7k+14=0\n-3k(k-2)-7(k-2)=0\n-(3k+7)(k-2)=0\nk=-(7)/(3) \vee k=2$

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Graph the system of inequalities presented here on your own paper, then use your graph to answer the following questions:

y > −4x − 1
y is less than 3 over 2 times x minus 1

Part A: Describe the graph of the system, including shading and the types of lines graphed. Provide a description of the solution area.

Part B: Is the point (−1, −1) included in the solution area for the system? Justify your answer mathematically.

Answers

A) the graph is a linear and the slope is negative and positive .

B) (-1,-1) is not a solution because that point does not appear in the shaded region, therefore there is no solution.

A quadrilateral with 90 degree angles is shown. The lengths of the left and right sides are 10 and 2 y + 4. The lengths of the top and bottom sides are 18 and 10 + 2 x.What is the value of x?

Answers

Answer:

x=3

Step-by-step explanation:

2 times 3 plus 4 equals 10

Answer:

x=3

Step-by-step explanation:

Which statement about vertical angles is false?A. They are formed by two intersecting lines
B. They share a common vertex
C. They share a common side
D. They are congruent