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Alexxandr [17]
3 years ago
14

Which word problem can best be represented by the fraction 5-8

Mathematics
1 answer:
FrozenT [24]3 years ago
3 0
The answer would be c because 1-5-7 is the same as 5-8
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What is the range of h (x)
Svet_ta [14]

The range of the function is the set of all possible outputs, that is, the set of all values obtained by applying the function to elements of the domain. So the set of all values which can be obtained by applying h(x) to an element of its domain is {−4,0,5,60} , and thus that is the range of h(x) .
6 0
3 years ago
Determine the value of X for the drawing below. ​
ASHA 777 [7]

Answer: x = 1

Step-by-step explanation: In short, 0 is the only number such that for any number x, x + 0 = x. ... Well, it's the only number which can be multiplied by any other number without changing that other number. In short, the multiplicative identity is the number 1, because for any other number x, 1*x = x.

5 0
2 years ago
Read 2 more answers
If the sum of three consecutive numbers is 852, what is the middle one?
denis-greek [22]

Divide the total by the quantity of numbers:

852/3 = 284

Because there are an odd quantity of numbers (3) the answer is the middle number.

The answer is 284

5 0
3 years ago
Again answer as many as you can
Keith_Richards [23]
For #3, it's 5.4 because you divide the total amount of money made by the amount of money earned per hour. 

49.95/9.25=5.4 hours
7 0
3 years ago
Read 2 more answers
For each of the following functions, determine if they are injective. Also determine if theyare surjective. Also determine if th
timurjin [86]

Answer:

Check below

Step-by-step explanation:

1. Definition for intervals

(a,b)=\left \{ x\in\Re : a

2. Functions

1) \Re \rightarrow \Re \\ f(x)=x

Let's perform graph tests.

That's an one to one, injective function. Look how any horizontal line touches that only once. Also, It's a surjective and a bijective one.

2)\Re\geq0\rightarrow\Re , f(x)=x+1\\

Injective, surjective and bijective.

Injective: a horizontal line crosses the graph in one point.

3)f:\Re\geq 0\rightarrow\Re, f(x) = cos(x)

The cosine function is not injective, bijective nor surjective.

4)f:\Re\rightarrow\Re \:f(x)=ex

Since e, is euler number it's a constant. It's also injective, surjective and bijective.

5)  Quite unclear format

6) \:f:\Re\rightarrow(0,\infty), f(x) =ex

Despite the Restriction for the CoDomain, the function remains injective, surjective and therefore bijective.

7) f:\Re\geq 0 \rightarrow  \Re\geq0, f(x) =x^{4}

Not injective nor surjective therefore not bijective too.

8).f:\{-1,2,-3\}}\rightarrow \{1,4,9\}, f(x) =x^{2}

f(-1)=1, f(2)=4, f(-3)=9

Injective (one to one), Surjective,  and Bijective.

10) f:\Re\geq 0\rightarrow [-1,1], f(x)= cos(x)\\-1=cos(x) \therefore x=\pi,3\pi,5\pi,etc.

Surjective.

11.f:R\geq 0[-1,1], f(x) = 0\\

Surjective

12.f: US Citizens→Z, f(x) = the SSN of x.

General function

13.f: US Zip Codes→US States, f(x) = The state that x belongs to.

Surjective

7 0
3 years ago
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