Aphysics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problems, in how many ways can the questions and problems be chosen?

720 ways

Step-by-step explanation:

Generally, combination is expressed as;

The question consists of 9 multiple-choice questions and examinee must answer 7 of the multiple-choice questions.

                                   ⇒ ⁹C₇

                                        = 36

The question consists of 6 open-ended problems and examinee must answer 3 of the open-ended problems.

                                    ⇒ ⁶C₃

                                         = 20

Combining the two combinations to determine the number of ways the questions and problems be chosen if an examinee must answer 7 of the multiple-choice questions and 3 of the open-ended problem.

                                        ⁹C₇ × ⁶C₃

                                       = 36 × 20

                                       = 720 ways  

30

step-by-step explanation:

idk

i hope i

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it