Type the correct answer in the box. Figure 1 is similar to figure 2. The height of the cone in figure 2 is centimeters.

9 Squared if equal to?

maria [59]

Answer: 81

Step-by-step explanation: 9 times 9 = 81

8 0

2 years ago

Read 2 more answers

Help me solve thank you will give 20 points

laila [671]

Answer:

Natalie bought 500 apples at $0.40 each, then she pays $0.40 500 times, this means that the total cost of the 500 apples is:

Cost = 500*$0.40 = $200

Now she threw away n apples from the 500 apples, then the number of apples that she has now is:

apples = 500 - n

And she sells the remaining apples for $0.70 each.

a) The amount that she gets by selling the apples is:

Revenue = (500 - n)*$0.70

b) We know that she did not make a loss, then the revenue must be larger than the cost, this means that:

cost ≤ revenue

$200 ≤ (500 - n)*$0.70

c) We need to solve the inequality for n.

$200 ≤ (500 - n)*$0.70

$200/$0.70 ≤ (500 - n)

285.7 ≤ 500 - n

n + 285.7 ≤ 500

n ≤ 500 - 285.7

n ≤ 214.3

Then the maximum value of n must be equal or smaller than 214.3

And n is a whole number, then we can conclude that the maximum number of rotten apples can be 214.

8 0

2 years ago

Kat is painting the edge of a triangular stage prop with reflective orange paint. The lengths of the edges of the triangle are (

vfiekz [6]

We are given with three lengths of a triangle expressed in terms and variables: (3x – 4) feet, (x^2 – 1) feet, and (2x^2 – 15) feet. The perimeter of the triangle is equal to the sum of the three sides of the triangle. In this case, the sum is 3x^2 + 3x -20. When x is equal to 4, we substitute <span>3*16 + 3*4 -20 equal to 40 feet.</span>

4 0

3 years ago

List the factors of 48 and 54

sergeinik [125]

48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

54: 1, 2, 3, 6, 9, 18, 27, 54

gcf: 6

4 0

3 years ago

Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

Svetlanka [38]

Answer:

(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.

5 0

2 years ago