What is 9,380 divided by 31

zzz [600]

Answer:

C

Step-by-step explanation:

A unit cube is a cube with equal dimensions of a single unit. This eliminates A, B, and D. C is the correct answer. Hope this helped ^^ Good luck

6 0

3 years ago

Read 2 more answers

What is the total number of women from the

Leviafan [203]

36 women have served.

5 0

3 years ago

24. Rectangle A is 16 feet wide and is 8 feet longer than Rectangle B which is 14 feet wide.

Galina-37 [17]

Answer:

The Area of rectangle A is 576 feet² .

Step-by-step explanation:

Given as :

The width of rectangle A = 16 feet

The Length of rectangle A = 8 feet + The length of rectangle B

The width of rectangle B = 14 feet

Let The length of rectangle B = L feet

So, The Length of rectangle A = 8 feet + L feet

The perimeter of rectangle A + The perimeter of rectangle B = 156 feet

So, 2 × ( Length A + width A ) + 2 × ( Length B + width B ) = 156 feet

Or, 2 × ( 8 + L + 16 ) + 2 × ( L + 14 ) = 156 feet

Or, 2 ×( 24 + L ) + 2 × ( L + 14 ) = 156 feet

Or, 48 + 2 L + 2 L + 28 = 156

Or, 76 L + 4 L = 156

So, 4 L = 156 - 76

Or, 4 L = 80

∴ L = $\frac{80}{4}$ = 20 feet

So , The Length of rectangle A = 8 feet + 28 feet = 36 feet

And The width of rectangle A = 16 feet

So, Area of rectangle A = Length of rectangle A × width of rectangle A

I.e Area of rectangle A = 36 × 16 = 576 feet²

Hence The Area of rectangle A is 576 feet² . Answer

4 0

3 years ago

Alexa has some dimes and some quarters. She has a minimum of 28 coins worth a maximum of $5.05 combined. If Alexa has 17 quarter

Lerok [7]

Answer:

There are no possible solutions

Step-by-step explanation:

Given:

Minimum number of coins Alexa has =28 coins

Number of quarters Alexa has = 17 quarters

Total amount in coins = $5.05 = 505 cents

Let number of dimes be = $x$ coins

So we can have two inequalities.

1) Total number of coins

$x+17\geq28\\$ [Since minimum number of coins=28]

2) Total value of coins

$10x+(25)(17)\leq505$ [As 1 dime=10 cents and 1 quarter=25 cents]

$10x+425\leq505$

Solving inequality (1)

Subtracting both sides by 17.

$x+17-17\geq28-17$

∴ $x\geq 11$

Solving inequality (2)

Subtracting both sides by 425.

$10x+425-425\leq505-425$

$10x\leq 80$

Dividing both sides by 10.

$\frac{10x}{10}\leq \frac{80}{10}$

∴ $x\leq 8$

On combining both solutions

$x\geq 11$ and $x\leq 8$ ,

we see that there are no possible solutions as number of dimes cannot be ≥11 and ≤8 at the same time.

7 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