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3 years ago

Find the rule, and type the missing number in the sequence. 920, 855, 790, ____

Mathematics

2 answers:

xenn [34]3 years ago

7 0

It’s “725”, just subtract it by 65..

Genrish500 [490]3 years ago

4 0

It’s 725,just by subtracting it by 65...

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3. Crystal has decided to drive her new car on the west coast from California to Washington, which is about 930

antoniya [11.8K]

Answer:

The correct option is the first graph (Yes)

Step-by-step explanation:

The given function is W(d) = 930/d

Where;

W(d) = The number of days it would take to get to Washington

d = The distance travelled each day in miles

We note that the given graphs have number of days in the y-axis and the miles in the x-axis, and that in the function, the distance Crystal travels each day is constant, d, while the 930 is the total distance from California to Washington, while W(d) is the time

Therefore, the variables for crystal are W(d) and the part of 930 miles completed, while the constant is the speed, d = miles/day, which gives;

W(d) = 930/d

W(d) ∝ Part of 930 miles completed × Constant

∴ W(d) which is the time is directly proportional to the distance

The graph which represents the number of days, W(d) it will take Crystal to get to Washington (miles completed) as a function of distance traveled per day, d, which is constant is a straight line graph which is the first graph

7 0

3 years ago

Which fraction represents the shaded part of the grid?

Alexxandr [17]

Answer:

30%

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Plz help ASAP I need help

Over [174]

Hotel a cause 40 x 4 = 160 which is under 175

6 0

3 years ago

Which equation matches the graph? PLEASE HELP MEEEE

Rama09 [41]

Answer: Y = 3x - 2

Work: Okay, let's take this step by step. Since the Y - Intercept is -2, put a dot on -2 on the (Y) Line. It also just so happens that the line goes through -2. Now with 3x. 3x is the same thing as 3x / 1. Since three is on the top, you go up three from -2, and then move over one spot because of the 1 (If it was a negative one, you go the opposite direction) Alright, now you see that the line also passes through that point as well, so it has to be y = 3x -2

I hope this helps, and Happy Holidays! :)

8 0

3 years ago

For x, y ∈ R we write x ∼ y if x − y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1)

vodomira [7]

Answer:

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

The relation being reflexive (∀a∈R, a∼a)
The relation being symmetric (∀a,b∈R, a∼b⇒b∼a)
The relation being transitive (∀a,b,c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it´s known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a,b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, -k ∈ Z

b∼a, ∀a,b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a,b,c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k,l ∈ Z

(a-b)+(b-c)=k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a,b,c ∈ R

∼ is Transitive by definition

We´ve shown that ∼ is an equivalence relation on R.

B. Now we have to show that there´s a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F(x)=[x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x,y∈[0,1), F(x)=F(y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F(x)=b):

F is injective:

let x,y ∈ [0,1) and suppose F(x)=F(y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x,y ∈ [0,1), then k must be 0. If it isn´t, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x,y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let´s find x such as x ∈ [0,1) and F(x)=[b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let´s show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F(r)=[b]

∼ is surjective

Then F maps [0,1) into C, i.e [0,1) is a set of representatives for the set of the equivalence classes.

4 0

3 years ago