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Mathematics 12 Online
OpenStudy (anonymous):

04.06 Quiz for algebra 2 help? If someone already has it? FAN AND MEDAL

OpenStudy (anonymous):

ou have been invited to a fancy dinner party celebrating your hard work in Algebra 2 so far. The distinguished guests come from various aspects of math disciplines like professors, engineers, and financial analysts. A mysterious box is delivered to the dinner party you are attending. The label on the box says that the volume of a box is the function f(x) = x3 + 3x2 – 10x – 24. To open the box, you need to identify the correct factors of f(x). Partygoers offer up solutions, and it is your job to find the right ones. Their suggestions are: (x – 1) (x + 2) (x – 3) (x + 4) (x + 6) (x – 12) List the correct factors. Then justify your selections with complete sentences. Three partygoers are in the corner of the ballroom having an intense argument. You walk over to settle the debate. They are discussing a function g(x). You take out your notepad and jot down their statements. Professor McCoy: She says that 2 is a zero of g(x) because long division with (x + 2) results in a remainder of 0. Ms. Guerra: She says that 2 is a zero of g(x) because g(2) = 0. Mr. Romano: He says that 2 is a zero of g(x) because synthetic division with 2 results in a remainder of 0. Correct the reasoning of any inaccurate reasoning by the partygoers in full and complete sentences. Make sure you reference any theorems that support your justifications. Dr. Collier summons you over to his table. He wants to demonstrate the graph of a fourth-degree polynomial function, but the batteries in his graphing calculator have run out of juice. Explain to Dr. Collier how to create a rough sketch of a graph of a fourth-degree polynomial function. Mrs. Collins is at the table with you and states that the fourth-degree graphs she has seen have 4 real zeros. She asks you if it is possible to create a fourth-degree polynomial with only 2 real zeros. Demonstrate how to do this and explain your steps.

jigglypuff314 (jigglypuff314):

Sorry but helpers are not allowed to help on quizes or tests, that would be considered cheating :/

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