Help please :) Thank you

Please help me branniest

Ksju [112]

There's no solution, as the shaded areas don't overlap

5 0

3 years ago

Tickets to a play are priced as follows: adult-£12 and child:£6

Darina [25.2K]

Cheryl has to pay a total charge of £100.98.

Step-by-step explanation:

Step 1; A ticket for an adult costs £12 and a ticket for a child costs £6. So for 6 adults and 3 children, the ticket price is

Ticket price= 6 × £12 + 3 × £6 = £72 + £18 = £90.

So the ticket prices alone are £90. A booking fee of 10% is added to this £90.

Booking fee = 10% of £90 = £9.

Ticket price inclusive of booking fee = £90 + £9 = £99.

Step 2; Cheryl pays by credit card so an additional 2% is added to the final price i.e. 2% is added to the £99.

Credit card charges = 2% × £99 = £1.98

Final booking fee = £99 + £1.98 = £100.98.

So buying the tickets with credit card, Cheryl pays £100.98 to buy the tickets.

7 0

3 years ago

Multiply then write the product in the simplest form <br> 3/1

Ganezh [65]

Answer:

9

Step-by-step explanation:

3/1*3/1=

3*3= 9

Answer: 9

9 is the most simplified version.

(I think this is what it means)

6 0

4 years ago

What is the algorithm for the inverse of f (x)?<br>

Juli2301 [7.4K]

Answer:

What is an inverse?

Recall that a number multiplied by its inverse equals 1. From basic arithmetic we know that:

The inverse of a number A is 1/A since A * 1/A = 1 (e.g. the inverse of 5 is 1/5)

All real numbers other than 0 have an inverse

Multiplying a number by the inverse of A is equivalent to dividing by A (e.g. 10/5 is the same as 10* 1/5)

What is a modular inverse?

In modular arithmetic we do not have a division operation. However, we do have modular inverses.

The modular inverse of A (mod C) is A^-1

(A * A^-1) ≡ 1 (mod C) or equivalently (A * A^-1) mod C = 1

Only the numbers coprime to C (numbers that share no prime factors with C) have a modular inverse (mod C)

Step-by-step explanation:

Please check image.

3 0

3 years ago

A subset $S \subseteq \mathbb{R}$ is called open if for every $x \in S$, there exists a real number $\epsilon > 0$ such that

const2013 [10]

Answer:

Step-by-step explanation:

REcall that given sets S,T if we want to prove that $S\subseteqT$ , then we need to prove that for all x that is in S, it is in T.

a) Let (a,b) be a non empty interval and $x\in (a,b)$ . Then a<x <b. Let $\varepsilon = \min{\min\{b-x, x-a\}}{2}$ Consider $y \in (x-\varepsilon,x+\varepsilon)$ , then

$y$ and

$y>x-\varepsilon>x-(x-a) = a$ .

Then $y\in (a,b)$ . Hence, (a,b) is open.

Consider the complement of [a,b] (i.e $(a,b)^c$ ).

Then, it is beyond the scope of this answer that

$(a,b)^c = (-\infty,a) \cup (b,\infty)$ .

Suppose that $x\in (a,b)^c$ and without loss of generality, suppose that x < a (The same technique applies when x>b). Take $\varepsilon = \frac{a-x}{2}$ and consider $y \in (x-\varepsilon,x+\varepsilon)$ . Then

$y$

Then y \in (-\infty,a). Applying the same argument when $x \in (b,\infty)$ we find that [a,b] is closed.

c) Let I be an arbitrary set of indexes and consider the family of open sets $\{A_i\}_{i\in I}. Let [tex]B = \bigcup_{i\in I}A_i$ . Let $x \in B$ . Then, by detinition there exists an index $i_0$ such that $x\in A_{i_0}$ . Since $A_{i_0}$ is open, there exists a positive epsilon such that $(x-\varepsilon,x+\varepsilon)\subseteq A_{i_0} \subseteq B$ . Hence, B is open.

d). Consider the following family of open intervals $A_n = (a-\frac{1}{n},b+\frac{1}{n})$ . Let $B = \bigcap_{n=1}^{\infty}A_n$ . It can be easily proven that

$B =[a,b]$ . Then, the intersection of open intervals doesn't need to be an open interval.

b) Note that for every $x \in \mathbb{R}$ and for every $\varepsilon>0$ we have that $(x-\varepsilon,x+\varepsilon)\subseteq \mathbb{R}$ . This means that $\mathbb{R}$ is open, and by definition, $\emptyset$ is closed.

Note that the definition of an open set is the following:

if for every $x \in S$ , there exists a real number $\epsilon > 0$ such that $(x-\epsilon,x \epsilon) \subseteq S$ . This means that if a set is not open, there exists an element x in the set S such that for a especific value of epsilon, the subset (x-epsilon, x + epsilon) is not a proper subset of S. Suppose that S is the empty set, and suppose that S is not open. This would imply, by the definition, that there exists an element in S that contradicts the definition of an open set. But, since S is the empty set, it is a contradiction that it has an element. Hence, it must be true that S (i.e the empty set) is open. Hence $\mathbb{R}$ is also closed, by definition. If you want to prove that this are the only sets that satisfy this property, you must prove that $\mathbb{R}$ is a connected set (this is a topic in topology)

6 0

3 years ago