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monitta
3 years ago
10

Five friends want to rent a go-cart track that costs $100. Each friend intends to contribute the same about. Susan used the five

-step problem-solving plan. After completing the solve step, how much money did she find that each friend will have to contribute to reach the goal? Enter you answer in the box.
Mathematics
2 answers:
elena-s [515]3 years ago
8 0

5 friends contribute x money to go to a cart that costs $100.
.
5x = 100
x = 100/5
x = 20
.
Each friend contributed $20 for the cart

Gemiola [76]3 years ago
6 0

20 dollars is ze answer


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Each mobile bags sold by Ravi’s marble company contains 4 purple marbles for every 7 orange marbles if the bag is 56 orange marb
Allisa [31]

Answer:

56/7 is 8 so 8x4 is 32 is is easy

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Y varies directly as x and inversely as the square of z. y equals 10 when x equals 36 and z equals 6. Find y when x equals 2 and
Ilia_Sergeevich [38]

Answer:

y = 20

Step-by-step explanation:

y =  \frac{kz}{x}  \\ 10 =  \frac{6k}{36}  \\ \frac{6k}{6}  = \frac{360}{6} \\ k= 60 \\  y =  \frac{4(60)}{2}  \\ y =  \frac{240}{2}  \\ y = 120

If this is right, then please make it the brainliest.

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3 years ago
Write an equation of the line that passes through (1,-2) and (-5,4)
denis23 [38]

Answer: x+y+1=0

Step-by-step explanation:

m=y2-y1/x2-x1

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y+2=-x+1

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7 0
2 years ago
Watch help video
polet [3.4K]

Answer:

is there any pic to refer the question?

Step-by-step explanation:

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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