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il63 [147K]
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
Mathematics
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

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enot [183]
4/10
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16/40
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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

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