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Alex
3 years ago
10

Distribute &combine 2 -(-x+5)

Mathematics
1 answer:
natita [175]3 years ago
8 0
2-(-x+5)
2+x-5
x-3
I decided to included the steps just in case you needed them

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Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations.
Nostrana [21]
x^3-27i=0
x^3=27i
x^3=27e^{i\pi/2}
x=\left(27e^{i\pi/2}\right)^{1/3}
x=3e^{i(\pi/2+2\pi k)/3}
x=3e^{i\pi(4k+1)/6}

where k\in\{0,1,2\}. This means you have

x=3e^{i\pi/6}=\dfrac32(\sqrt3+i)
x=3e^{i5\pi/6}=\dfrac32(-\sqrt3+i)
x=3e^{i9\pi/6}=-3i

as the solutions to the original equation.
4 0
3 years ago
Mike has 8 feet of rope . How many inches of rope does he have.
Ostrovityanka [42]
96 in. of rope.
8x12=96
4 0
3 years ago
Read 2 more answers
The average number of points a basketball team scored for
Ganezh [65]
Average is

(2x + X+6)/3=63
(2x+x+6)= 189
3x+6=189
3x=183 61

X=61
Scores for the game are: 61, 61, and 67
3 0
3 years ago
A fair coin is tossed three times in succession. The set of equally likely outcomes is StartSet HHH comma HHT comma HTH comma TH
liq [111]

The probability of getting exactly zero tails is 1/8.

Step-by-step explanation:

When a fair coin is tossed three time, the set of equally likely outcomes is:

S= {HHH,HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S)=8

Let A be the event that there are exactly zero tails which means there are all heads

Then

A= {HHH}

n(A)=1

P(A)=\frac{n(A)}{n(S)}\\=\frac{1}{8}

The probability of getting exactly zero tails is 1/8.

Keywords: Probability, Equally likely events

Learn more about probability at:

  • brainly.com/question/13219835
  • brainly.com/question/1836777

#LearnwithBrainly

5 0
2 years ago
3. Let A, B, C be sets and let ????: ???? → ???? and ????: ???? → ????be two functions. Prove or find a counterexample to each o
Fiesta28 [93]

Answer / Explanation

The question is incomplete. It can be found in search engines. However, kindly find the complete question below.

Question

(1) Give an example of functions f : A −→ B and g : B −→ C such that g ◦ f is injective but g is not  injective.

(2) Suppose that f : A −→ B and g : B −→ C are functions and that g ◦ f is surjective. Is it true  that f must be surjective? Is it true that g must be surjective? Justify your answers with either a  counterexample or a proof

Answer

(1) There are lots of correct answers. You can set A = {1}, B = {2, 3} and C = {4}. Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. Then g is not  injective (since both 2, 3 7→ 4) but g ◦ f is injective.  Here’s another correct answer using more familiar functions.

Let f : R≥0 −→ R be given by f(x) = √

x. Let g : R −→ R be given by g(x) = x , 2  . Then g is not  injective (since g(1) = g(−1)) but g ◦ f : R≥0 −→ R is injective since it sends x 7→ x.

NOTE: Lots of groups did some variant of the second example. I took off points if they didn’t  specify the domain and codomain though. Note that the codomain of f must equal the domain of

g for g ◦ f to make sense.

(2) Answer

Solution: There are two questions in this problem.

Must f be surjective? The answer is no. Indeed, let A = {1}, B = {2, 3} and C = {4}.  Then define f : A −→ B by f(1) = 2 and g : B −→ C by g(2) = 4 and g(3) = 4. We see that  g ◦ f : {1} −→ {4} is surjective (since 1 7→ 4) but f is certainly not surjective.  Must g be surjective? The answer is yes, here’s the proof. Suppose that c ∈ C is arbitrary (we  must find b ∈ B so that g(b) = c, at which point we will be done). Since g ◦ f is surjective, for the  c we have already fixed, there exists some a ∈ A such that c = (g ◦ f)(a) = g(f(a)). Let b := f(a).

Then g(b) = g(f(a)) = c and we have found our desired b.  Remark: It is good to compare the answer to this problem to the answer to the two problems

on the previous page.  The part of this problem most groups had the most issue with was the second. Everyone should  be comfortable with carefully proving a function is surjective by the time we get to the midterm.

3 0
3 years ago
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