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makvit [3.9K]
2 years ago
7

Lynn gave her nail technician a $5.10 tip. The tip was 15% of the cost of the manicure. Write an

Mathematics
1 answer:
satela [25.4K]2 years ago
6 0

Answer:

Step-by-step explanation:

b x .15 = 5.10

b=5.10/.15

b=34

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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
What is the square root of 383748
dybincka [34]
<span>619.473970397 here is the square root</span>
8 0
3 years ago
Read 2 more answers
Determine the independent variable (__________,y). what word goes in the blank space??
bulgar [2K]
In the blank the answer is x axis
5 0
3 years ago
Pls help I will give points
s2008m [1.1K]

Answer:

60

Step-by-step explanation:

5 0
3 years ago
1+-w2+9w and I need help cuz I’m on 76 and I’m sooo close help
Gnesinka [82]

\huge \boxed{\mathfrak{Question} \downarrow}

  • Simplify :- 1 + - w² + 9w.

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

\large \sf1 + - w ^ { 2 } + 9 w

Quadratic polynomial can be factored using the transformation \sf \: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where \sf x_{1} and x_{2} are the solutions of the quadratic equation \sf \: ax^{2}+bx+c=0.

\large \sf-w^{2}+9w+1=0

All equations of the form \sf\:ax^{2}+bx+c=0 can be solved using the quadratic formula: \sf\frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

\large \sf \: w=\frac{-9±\sqrt{9^{2}-4\left(-1\right)}}{2\left(-1\right)}  \\

Square 9.

\large \sf \: w=\frac{-9±\sqrt{81-4\left(-1\right)}}{2\left(-1\right)}  \\

Multiply -4 times -1.

\large \sf \: w=\frac{-9±\sqrt{81+4}}{2\left(-1\right)}  \\

Add 81 to 4.

\large \sf \: w=\frac{-9±\sqrt{85}}{2\left(-1\right)}  \\

Multiply 2 times -1.

\large \sf \: w=\frac{-9±\sqrt{85}}{-2}  \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is plus. Add -9 to \sf\sqrt{85}.

\large \sf \: w=\frac{\sqrt{85}-9}{-2}  \\

Divide -9+ \sf\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{9-\sqrt{85}}{2}} \\

Now solve the equation \sf\:w=\frac{-9±\sqrt{85}}{-2} when ± is minus. Subtract \sf\sqrt{85} from -9.

\large \sf \: w=\frac{-\sqrt{85}-9}{-2}  \\

Divide \sf-9-\sqrt{85} by -2.

\large \boxed{ \sf \: w=\frac{\sqrt{85}+9}{2}}  \\

Factor the original expression using \sf\:ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \sf\frac{9-\sqrt{85}}{2}for \sf\:x_{1} and \sf\frac{9+\sqrt{85}}{2} for \sf\:x_{2}.

\large \boxed{ \boxed {\mathfrak{-w^{2}+9w+1=-\left(w-\frac{9-\sqrt{85}}{2}\right)\left(w-\frac{\sqrt{85}+9}{2}\right) }}}

<h3>NOTE :-</h3>

Well, in the picture you inserted it says that it's 8th grade mathematics. So, I'm not sure if you have learned simplification with the help of biquadratic formula. So, if you want the answer simplified only according to like terms then your answer will be ⇨

\large \sf \: 1 + -  w {}^{2}  + 9w \\  =\large  \boxed{\bf \: 1 -  {w}^{2}   + 9w}

This cannot be further simplified as there are no more like terms (you can use the biquadratic formula if you've learned it.)

4 0
2 years ago
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