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2 years ago

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Elliot drew this model to solve the problem -3+4 but then realized he made a mistake. What addition problem is represented by El

liot's model?

1 answer:

denis23 [38]2 years ago

4 0

Answer: -3+4 so he made a mistake so it would be -7

Step-by-step explanation:

You might be interested in

Select the correct answer.

Nina [5.8K]

Answer:

D. $x^2 + 3x - 88$

Step-by-step explanation:

Area of rectangle = $length (l) * width (w)$

Length of rectangle = (x - 8)

Width = (x + 11)

Area of rectangle = $(x - 8)(x + 11)$

Expand the expression

$x(x + 11) -8(x + 11)$ (distributive property of multiplication)

$x^2 + 11x - 8x - 88$

Combine like terms

$x^2 + 3x - 88$

Expression for the area of the rectangle = $x^2 + 3x - 88$

7 0

2 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:

<h3>(a)</h3>

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.

<h3>(b)</h3>

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

Show all your work to receive full credit for this problem. Solve the system of equations using the elimination method. 2a + 3b

Sindrei [870]

Given equations are;<span>
2a + 3b = -1 ..................equation 1
3a + 5b = -2 ....................equation 2</span>

Now multiply equation 1 with (-3)

The equation will be;

-6a -9b = 3 …………………..equation 3

Now multiply equation 2 with (2)

The equation will be;

6a + 10b = -4 ……………..equation 4

Now add equation 3 and equation 4

-6a – 9b = 3

<span>6a + 10b = -4</span>

<span>------------------------------</span>

0a + b = -1

b = -1

Now put the value of b in equation 1

2a + 3(-1) = -1

2a -3 = -1

2a = -1+3

2a = 2

a=1

Thus the solution is (a,b) = (1,-1)

<span>
</span>

8 0

3 years ago

Lines m and n are parallel. Which angle has a measure of 110°?

grigory [225]

This might help you..need more information

6 0

2 years ago

anthony is solving the equation x^2-12x=16 by completing the square. what number should be added to both sides of the equation t

zimovet [89]

Answer:

add (12/2)^2 to both sides

Step-by-step explanation:

x^2-12x=16

x^2 -12x + (12/2)^2 = 16 + (12/2)^

x- (12/2)^2=52

(x-6)^2=52

Answers:

x= -2 root 3 +6

x= 2 root 3 +6

4 0

3 years ago