Binomial expansion (3x-5y)^3

I need help aaaaaaaaa

jek_recluse [69]

Answer:

the correct answer is d

7 0

1 year ago

Read 2 more answers

How do I solve the length of AC?

Crank

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∆ABC and ∆ADE are similar triangles.

Thus ;

$\frac{AD}{AC} = \frac{DE}{BC} = \frac{AE}{AB} \\$

So :

$\frac{5}{3 + 2x} = \frac{4}{8} \\$

$\frac{5}{3 + 2x} = \frac{1}{2} \\$

Inverse both sides

$\frac{3 + 2x}{5} = \frac{2}{1} \\$

$\frac{2x + 3}{5} = 2 \\$

Multiply sides by 5

$5 \times \frac{2x + 3}{5} = 5 \times 2 \\$

$2x + 3 = 10$

Thus ;

$AC = 2x + 3 = 10$

$AC = 10$

Done...

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7 0

2 years ago

Please help me!!!!!!!!

Bezzdna [24]

Answer:

(3·x + 6) + (7·x + 4) = 180 --> x = 17

3·17 + 6 = 57

Answer D is correct.

3 0

2 years ago

You get a job making $12 per hour. You do a great job and your boss says he is going to give you a 25% raise!!! What is your new

AURORKA [14]

Answer:

15

Step-by-step explanation:

10% of 12 =1.2

1.2*2= 2.4 (20%)

1.2/2= 0.6 (5%)

2.4+0.6= 3

12+3= $15

3 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

2 years ago