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tigry1 [53]
3 years ago
5

Simplify this expression.

Mathematics
2 answers:
sesenic [268]3 years ago
7 0

Answer:

C

Step-by-step explanation:

use distributive property and get 2x -8

do 2(10) and get 20

20 - 8

get 2x +12

frozen [14]3 years ago
6 0

Answer:

2x + 12

Step-by-step explanation:

2(10)+2(x+4)

distribute the 2

20 + 2x - 8

add/ subtract like terms

2x + 12

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Given that sin theta = 1/4, 0
GaryK [48]
Answer: cos(Θ) = (√15) / 4

Explanation:

The question states:

1) sin(Θ) = 1/4

2) 0 < Θ < π / 2

3) find cos(Θ)

This is how you solve it.

1) Use the fundamental identity (in this part I use α instead of Θ, just for facility of wirting the symbols, but they mean the same for the case).

(cos \alpha )^2 + (sin \alpha )^2 =1

2) From which you can find:

(cos \alpha )^2 = 1 - (sin \alpha )^2

3) Replace sin(α) with 1/4

=> (cos \alpha )^2 = 1 - (1/4)^2 = 1 - 1/16 = 15/16

=> cos \alpha =+/- \sqrt{15/16} = +/- (\sqrt{15} )/4

4) Given that the angle is in the first quadrant, you know that cosine is positive and the final answer is:

cos(Θ) = \sqrt{15} /4.

And that is the answer.
6 0
3 years ago
For f(x)= 3x+1 and g(x)= x^2-6, find (g/f)(x).
o-na [289]
D. The exclusion is important because we cannot divide by zero. Apart from that we are simply looking at a function. 
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3 years ago
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Dado un triángulo rectángulo cuyo perímetro es de 48 cm, la diferencia entre su lado mayor y su lado menor es de 8 cm y su área
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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
How do you add 1/15 + 1/6?
Brut [27]
you are going to need to find the common name of common denominator 4/1 over 4 plus 1 over 6 equal to look for the commentator of 15 and 6 what goes into 15 and 6 is going to 15 and 6 the common denominator will be actually 12 to 6 only go into 15 2 times cuz I can't get close to 15 than q2 so you have to common denominator now this is going to be both 1/12 + 1/12 is going to be either 2/24 or 2/12
6 0
3 years ago
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