Multiply the polynomials. (2x2 + 6x + 6)(3x – 2)

Mathematics

1 answer:

gavmur [86]3 years ago

6 0

Answer:

$6x^{3}+16x^{2} +6x-12$

Step-by-step explanation:

To calculate the result of that mutiplication you have to firs evaluate the first term of the second function multipying the whole first function, woudl be like this:

$(3x)(2x^{2}+6x+6)= 6x^{3}+18x^{2} +18x$

Now that you have this, you multiply the second term from the second function to obtain:

$(-2)(2x^{2}+6x+6)= -4x^{2}-12x-12$

Now you just have to add both results:

$6x^{3}+18x^{2} +18x$

+ $-4x^{2}-12x-12$

And the result of the sum would be:

$6x^{3}+16x^{2} +6x-12$

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If the area of a square is 100 cm2, what is the perimeter?

Vitek1552 [10]

Answer:

Step-by-step explanation:

To find the perimeter, first we need to find the side lengths.

If 100 cm is the area then we need to find what number squared is 100

10 x 10 = 100

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2 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$