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Artist 52 [7]
3 years ago
11

Mary's car can go 24 miles on one gallon of gasoline. However , her gas mileage can vary by 2 miles per gallon depending on wher

e she drives . Write an absolute value equation that you can use to find the minimum and maximum gas mileage.
Mathematics
1 answer:
lakkis [162]3 years ago
4 0

Answer: | x -24 |=2

Step-by-step explanation:

Given: Mary's car can go 24 miles on one gallon of gasoline.

her gas mileage can vary by 2 miles per gallon depending on where she drives .

i.e. either 2 miles added or subtracted in the  gas mileage.

i.e. -2< Gas milegae -24 <2

⇒  | Gas milegae -24 |=2

Let x represents the gas mileage.

Then,  | x -24 |=2 is the required absolute value equation that you can use to find the minimum and maximum gas mileage.

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At the start of a science experiment, the temperature is 10 degrees Celsius. If the temperature decreases by a constant rate of
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Answer:

-5 If it is one of your answers it will be -5

Step-by-step explanation:

If you take 10 and have 3 decrease every minute and did it for 5 minutes it will go 10   7    4    1   -2    -5 That is 5 minutes of the temperature decreasing by 3 I hope it is right.

 

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Triangle ABC is similar to triangle EFD. What scale factor is required to dilate triangle ABC so that it's image, A'B'C' , is co
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A ramp is to be built beside the steps to the campus library. Find the angle of elevation of the 23​-foot ​ramp, to the nearest
Usimov [2.4K]

Answer:

the angle of elevation is 12.56°

Step-by-step explanation:

the height of the ramp represents the opposite side and the length of the ramp the hypotenuse

we see that it has (angle, hypotenuse, opposite)

well to start we have to know the relationship between angles, legs and the hypotenuse

a: adjacent

o: opposite

h: hypotenuse

sin α = o/h

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tan α = o/a

we choose the one with opposite and hypotenuse

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the angle of elevation is 12.56°

3 0
3 years ago
Consider the function y = cos (x + pi/6). Which of the following is true?
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7 0
3 years ago
Read 2 more answers
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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