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3 years ago

What is the factored form of 18+12

Mathematics

2 answers:

Ulleksa [173]3 years ago

8 0

6(3+2)

Step-by-step explanation:

that xas factor of 18+12

lesantik [10]3 years ago

3 0

Answer:

30:)

Step-by-step explanation:

I think it's 30!!!!!

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=<br> 2<br> у<br> when x = 21<br> y=3* + 20

ehidna [41]

Answer:

Step-by-step explanation:

7 0

3 years ago

The ratio of chocolate chip cookies, snickerdoodles and oatmeal raisin cookies was 2:5:7. During the class party 12 chocolate ch

Schach [20]

Answer: 127 cookies

Step-by-step explanation:

1. If you look back to the text you will see that 12 is 1 in 1:3:4 and you multiply that by 2. 12 * 2 = 24.

2. Divide 24 by 3 and that gives you 8 than times it by 5 for it to give you 40

3. Divide 36 by 4 and that equals to 9 than multiply it by 7 which that gives you 63

4. Add 24, 40, and 63 for it to give you 127

Thank you for following these steps if it is correct and this is my first time so if I am incorrect I will look upon my mistake and correct and help many of you people who need help

4 0

3 years ago

Need help Will give you the brainliest

marshall27 [118]

Answer:

The missing area for both sides is 48ft²

the missing dimension is 12ft

Step-by-step explanation:

Add all the areas up

12+36+12+36=96

192-96=96

96÷2=48

48÷4=12

5 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

1+4=5<br> 2+5=12<br> 3+6=21<br> 8+11=

jek_recluse [69]

=12-5=7
=21-12=9
when u look at they have a diffrence of 2 so the answer will have a diffrence of 2 =11

8 0

3 years ago