Ask question

Ask question

All categories

2 years ago

14

BRAINLIEST TO FIRST RIGHT ANSWER The coefficients of the first three terms in the expansion of (x – y) 4 are a) 1, –4, –6 b) 1,

–4, 6 c) 1, 4, 6 d) 1, 3, 5

1 answer:

Zigmanuir [339]2 years ago

7 0

Answer:

b) 1, -4, 6

Step-by-step explanation:

(x-y)^4=

(x-y)(x-y)(x-y)(x-y)=

(x^2-xy-xy-y^2)(x^2-xy-xy-y^2)=

(x^2-2xy-y^2)(x^2-2xy-y^2)=

x^4-4x^3y+6x^2y^2-4xy^3+y^4

You might be interested in

What are 5 fractions that can simplify to 2/5

enot [183]

4/10
8/20
16/40
32/80
64/160

6 0

3 years ago

Read 2 more answers

Find the value of each expression. Show your work.

Verdich [7]

Answer:

$A) (1.2)^{2} = 1.44$

$B)2^ 3+17-(3\times 4) = 13$

$C)\frac{(9)^2}{(3)^3} = 3$

Step-by-step explanation:

Here, the given expressions are:

A) $(1.2)^{2}$

Solving this, we get

$(1.2)^{2} = 1.2 \times 1.2 = 1.44$

⇒ $(1.2)^{2} = 1.44$

B) $2^ 3+17-(3\times 4)$

Now, solving this, we get

$2^ 3+17-(3\times 4) = (2 \times 2 \times2) + (17 -12)\\=8 + 5 = 13$

⇒ $2^ 3+17-(3\times 4) = 13$

C) $\frac{(9)^2}{(3)^3}$

Simplifying this, we get

$\frac{(9)^2}{(3)^3} = \frac{(9\times9)}{(3 \times 3\times3)} = 3$

⇒ $\frac{(9)^2}{(3)^3} = 3$

4 0

3 years ago

You are given a list of students names and their test scores. Design a pseudocode that does the following:(a) Calculates the ave

Rashid [163]

Answer:

List_of_students = {}; //Structure array that contains the scores of students and Names

Number_of_students = size(List_of_students);

Average_score ; //Variable Stores average score

Below_Average = []; //That is initially filled with zeros

Highest_score ; //Initialize variable to store highest score

for ( i = 0 ; i < Number_of_students; i = i + 1)

{

Average_score = Average_score + List_of_students(i);

if ( i = 15)

{

Average_score = Average_score / Number_of_students;

}

}

for ( j = 0; j < Number_of_students ; i = i + 1)

{

if ( List_of_students(j) < Average_score)

{

print( "student %name got %score , List_of_students(j), List_of_students(j) )

}

}

Compare_Score = 0; //Variable to compare score

for ( k = 0; K < Number_of_students; k = k+1)

{

if ( List_of_students(k) => Compare_Score)

{

Compare_Score = List_of_students(k);

}

}

Highest_Score = Compare_Score;

Step-by-step explanation:

4 0

3 years ago

Find the probability that the mean annual preciptiation will be between 32 and 34 inches. variable is normally distributed

liq [111]

Supposing, for the sake of illustration, that the mean is 31.2 and the std. dev. is 1.9.

This probability can be calculated by finding z-scores and their corresponding areas under the std. normal curve.
34 in - 31.2 in
The area under this curve to the left of z = -------------------- = 1.47 (for 34 in)
1.9
32 in - 31.2 in
and that to the left of 32 in is z = ---------------------- = 0.421
1.9

Know how to use a table of z-scores to find these two areas? If not, let me know and I'll go over that with you.

My TI-83 calculator provided the following result:

normalcdf(32, 34, 31.2, 1.9) = 0.267 (answer to this sample problem)

5 0

2 years ago

(20 POINTS AND BRAINLIEST FOR BEST)

Bingel [31]

Answer:

a. 8.5a + 38

b. 6+h+x

3 0

2 years ago