1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Leona [35]
3 years ago
14

Divide £50 in to the ratio 1 :4

Mathematics
1 answer:
Charra [1.4K]3 years ago
5 0
50 divided by 4 is 12.5 so the ratio 1:4 ratio will be 12.5/50
You might be interested in
Need help adding these<br><br> √54 + √24
Gennadij [26K]

Answer:

12.247 Hope I helped

Step-by-step explanation:

\sqrt{54}+\sqrt{24}  \\7.348+4.899\\12.247

8 0
3 years ago
Read 2 more answers
Ann took a taxi home from the airport. The taxi fare was $2.10 per mile, and she gave the driver a tip of $5. Ann paid a total o
adell [148]

Answer:

$2.10x + 5= $49.10

x=21

Step-by-step explanation:

the x is used for a variable that changes and in this case the mph changes which lead x being the amount of miles that they had drove which at the end ends up multi. with the $2.10 . as for the 5 it was an initial as in it doesn't change.

for the second part

you subtract 5 from each side so you have the total of 2.10x= 44.10 then u divide 2.10 from each side and that will give you x = 21

3 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
Write a polynomial function of minimum degree in standard form with real coefficients whose zeros and their multiplicities inclu
Softa [21]

<u>Answer-</u>

<em>The polynomial function is,</em>

y=x^4-10x^3+38x^2-64x+40

<u>Solution-</u>

The zeros of the polynomial are 2 and (3+i). Root 2 has multiplicity of 2 and (3+i) has multiplicity of 1

The general form of the equation will be,

\Rightarrow y=(x-(2))^2(x-(3+i))(x-(3-i))   ( ∵ (3-i) is the conjugate of (3+i) )

\Rightarrow y=(x-2)^2(x-3-i)(x-3+i)

\Rightarrow y=(x^2-4x+4)((x-3)-i)((x-3)+i)

\Rightarrow y=(x^2-4x+4)((x-3)^2-i^2)

\Rightarrow y=(x^2-4x+4)((x^2-6x+9)+1)

\Rightarrow y=(x^2-4x+4)(x^2-6x+10)

\Rightarrow y=x^2x^2-6x^2x+10x^2-4x^2x+4\cdot \:6xx-4\cdot \:10x+4x^2-4\cdot \:6x+4\cdot \:10

\Rightarrow y=x^4-10x^3+14x^2+24x^2-40x-24x+40

\Rightarrow y=x^4-10x^3+38x^2-64x+40

Therefore, this is the required polynomial function.


4 0
3 years ago
After substituting, what is the first operation performed when evaluating 5 + (9 x minus 3) divide 2 for x = 5? addition multipl
ludmilkaskok [199]

Answer:

<h2>Multiplication.</h2>

Step-by-step explanation:

The given expression is

5+(9x-3) \div 2

For x=5, we have

5+(9(5)-3) \div 2

Notice that the first operation we need to do is multiplication, because that's the right order of operations when we solve these kind of expressions.

5+(45-3) \div 2

Then, we subtract

5+42 \div 2

Now, we divide

5+21

Finally, we sum

26

Therefore, the right answer is multiplication.

4 0
3 years ago
Read 2 more answers
Other questions:
  • LCD of 1/5 and 9/10?
    12·1 answer
  • What is the greatest common factor of 180 and 240
    9·2 answers
  • A student earns a flat rate of $30/day planting trees, plus $0.50/tree planted. Create a table of values and scatter plot showin
    7·1 answer
  • How do you solve this equation 1/3(x+1)+2x=2.
    15·1 answer
  • Please help 3(6x−2)=−2(4x−9)+2
    13·2 answers
  • The steps below describe the construction of a line AG that is parallel to segment PQ and passes through a point A outside of PQ
    13·2 answers
  • Karen is knitting quilts for her family and friends as holiday gifts. As she knits the quilts she becomes more proficient each w
    11·1 answer
  • Select three ratios that are equivalent to 4:18
    10·1 answer
  • Simplify<br> √(8+2√15)<br><br> The decimal answer won't work.
    8·1 answer
  • 7 + 6d + 5f use d = 3 and f = 9
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!