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2 years ago

Calculate the coefficients of the first four terms of the binomial expansion for the binomial (x+y)^28

1 answer:

8 0

Consider the following expanded powers of (a + b)n, where a + b is any binomial and n is a whole number. Look for patterns.

Each expansion is a polynomial. There are some patterns to be noted.

1. There is one more term than the power of the exponent, n. That is, there are terms in the expansion of (a + b)n.

2. In each term, the sum of the exponents is n, the power to which the binomial is raised.

3. The exponents of a start with n, the power of the binomial, and decrease to 0. The last term has no factor of a. The first term has no factor of b, so powers of b start with 0 and increase to n.

4. The coefficients start at 1 and increase through certain values about "half"-way and then decrease through these same values back to 1.

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Point b is between points a and C. AB=2x+5 and BC=5x-1. What is the length of AB

telo118 [61]

Answer: AB = 9

Step-by-step explanation:

Since B is the midpoint, AB and BC must be congruent (since theyre both half of the segment). The equation is:

2x + 5 = 5x - 1

This means x = 2, meaning AB is 9.

7 0

2 years ago

Use matrices to solve the system of equations if possible. Use Gaussian elimination with back substitution or gauss Jordan elimi

CaHeK987 [17]

In matrix form, the system is given by

$\begin{bmatrix} -1 & 1 & -1 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -20 \\ 29 \\ 29 \end{bmatrix}$

I'll use G-J elimination. Consider the augmented matrix

$\left[ \begin{array}{ccc|c} -1 & 1 & -1 & -20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Multiply through row 1 by -1.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 2 & -1 & 1 & 29 \\ 3 & 2 & 1 & 29 \end{array} \right]$

• Eliminate the entries in the first column of the second and third rows. Combine -2 (row 1) with row 2, and -3 (row 1) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 5 & -2 & -31 \end{array} \right]$

• Eliminate the entry in the second column of the third row. Combine -5 (row 2) with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 3 & 24 \end{array} \right]$

• Multiply row 3 by 1/3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entry in the third column of the second row. Combine row 2 with row 3.

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

• Eliminate the entries in the second and third columns of the first row. Combine row 1 with row 2 and -1 (row 3).

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 9 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 8 \end{array} \right]$

Then the solution to the system is

$\boxed{x=9, y=-3, z=8}$

If you want to use G elimination and substitution, you'd stop at the step with the augmented matrix

$\left[ \begin{array}{ccc|c} 1 & -1 & 1 & 20 \\ 0 & 1 & -1 & -11 \\ 0 & 0 & 1 & 8 \end{array} \right]$

The third row tells us that $z=8$ . Then in the second row,

$y-z = -11 \implies y=-11 + 8 = -3$

and in the first row,

$x-y+z=20 \implies x=20 + (-3) - 8 = 9$

5 0

1 year ago

Anna is five years old. my age is six years more than thrice her age. how old am i

Hoochie [10]

Answer:

21 years old

Step-by-step explanation:

Anna = 5

You = 6 + (5)3

Equation =

(5)3 + 6

15 + 6 = 21

You are 21

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

4 0

2 years ago

Answer choices A. -4 B. 5.8 C.4 D.8

ankoles [38]

Answer:

the answer is C (4)

8 0

3 years ago

Write 358.41 x 10^5 as a standard form. 10 minutes left plzz helpppp

myrzilka [38]

Answer: 35841000

Step-by-step explanation: Well basically 10^5 is 10 followed by 4 zeros (100000). Knowing this multiply 358.41*100000 which is 35841000.

I think this is correct but I may have done something wrong

3 0

2 years ago