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Mathematics

1 answer:

jekas [21]3 years ago

7 0

$(r+s)(x)=r(x)+s(x)\\\\(r\cdot s)(x)=r(x)\cdot s(x)\\\\(r-s)(x)=r(x)-s(x)\\\\r(x)=x-4,\ s(x)=2x+2.\ \text{Substitute:}\\\\(r+s)(x)=x-4+2x+2=(x+2x)+(-4+2)=\boxed{3x-2}\\\\(r\cdot s)(x)=(x-4)(2x+2)=(x)(2x)+(x)(2)+(-4)(2x)+(-4)(2)\\\\=2x^2+2x-8x-8=\boxed{2x^2-6x-8}\\\\(r-s)(x)=x-4-(2x+2)=x-4-2x-2=(x-2x)+(-4-2)\\\\=\boxed{-x-6}$

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At a local University, it is reported that 81% of students own a wireless device. If 5 students are selected at random, what is

Sphinxa [80]

Answer:

34.87% probability that all 5 have a wireless device

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they own a wireless device, or they do not. The probability of a student owning a wireless device is independent from other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$