All categories

3 years ago

1) What is the equation of the line whose y-intercept is 3 and slope is 1?

Mathematics

2 answers:

s2008m [1.1K]3 years ago

5 0

$The\ slope-intercept\ form:y=mx+b\\\\m\ is\ slope\\b\ is\ y-intercept\\\\m=1;\ b=3\ then\ y=1x+3\Rightarrow y=x+3$

Studentka2010 [4]3 years ago

4 0

$y-intercept : \ b= 3 \\ slope : \ m=1 \\ \\ the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\ \\ y=1\cdot x + 3 \\ \\y=x+3$

You might be interested in

Kellie is playing with a toy building block that has the shape of a rectangular prism. What is the volume of the toy building bl

Nikolay [14]

The picture in the attached figure

we know that
volume of <span>a rectangular prism=L*W*H
where
L=4 in
W=5 in
H=10 in
so
</span>volume of a rectangular prism=4*5*10----> 200 in³
<span>
the answer is
200 in</span>³<span>

</span>

4 0

3 years ago

I need help I am so smart totally.

Contact [7]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

If f(x)=3x+2 and g(x)=x2-x find each value. 14. g(2)-2

Katena32 [7]

Value 14 so it must be 55 or something

4 0

3 years ago

Help me plzzzzzzzzzzzzzzzzzzz

lubasha [3.4K]

Answer:

Step-by-step explanation:

irdk sorry if its wrong :(

5 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:

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

3 years ago