Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

108,789 users

Algebra 2 homework help please!?

0 votes
1. Expand (2x+3)^5

2. (X+15)/ (X+6) ≥2
asked May 30, 2013 in ALGEBRA 2 by dkinz Apprentice

5 Answers

0 votes

1.The expression is (2x+3)^5

Now we use the formula image

Let x = 2x, y = 3 and n = 5

Apply formula

image

(The values of image)

image

(The values of image)

image

The expansion of image.

answered May 30, 2013 by dozey Mentor
0 votes

2). The inequality is (X+15)/ (X+6) ≥2

Multiply each side by (X + 6)

{(X+15)/ (X+6)}*(X + 6) ≥2(X + 6)

(X+15) ≥2(X+6)                                          (Simplify)

(X+15) ≥2X+12                                          (Multiply)

Subtract X from each side

X+15 - X ≥2X+12 - X

15 ≥X+12                                                   (Simplify)

Subtract 12 from each side

15 - 12 ≥X+12 - 12

3 ≥X                                                           (Simplify)

The inequality solution set is {X / X ≤ 3}.

answered May 30, 2013 by dozey Mentor

Solution of the inequality (X+15)/ (X+6) ≥2 is  -6 < ≤ 3.

0 votes

1.open parentheses 2 x plus 3 close parentheses to the power of 5 equal 5 subscript c subscript 0 end subscript open parentheses 2 x close parentheses to the power of 5 minus 0 end exponent open parentheses 3 close parentheses to the power of 0 plus 5 subscript c subscript 1 end subscript open parentheses 2 x close parentheses to the power of 5 minus 1 end exponent open parentheses 3 close parentheses plus 5 subscript c subscript 2 end subscript open parentheses 2 x close parentheses to the power of 5 minus 2 end exponent open parentheses 3 close parentheses to the power of 2 plus 5 subscript c subscript 3 end subscript open parentheses 2 x close parentheses to the power of 5 minus 3 end exponent open parentheses 3 close parentheses to the power of 3 plus 5 subscript c subscript 4 end subscript open parentheses 2 x close parentheses to the power of 5 minus 4 end exponent open parentheses 3 close parentheses to the power of 4 plus 5 subscript c subscript 5 end subscript open parentheses 2 x close parentheses to the power of 5 minus 5 end exponent open parentheses 3 close parentheses to the power of 5 space space space space left parenthesis A p p l y space b i o n o m i a l space t h e o r e m space right parenthesis

space space space space space space space space space space space space space space space space space space equal 1 open parentheses 2 x close parentheses to the power of 5 open parentheses 1 close parentheses plus 5 open parentheses 2 x close parentheses to the power of 4 open parentheses 3 close parentheses plus 10 open parentheses 2 x close parentheses to the power of 3 open parentheses 9 close parentheses plus 10 open parentheses 2 x close parentheses to the power of 2 open parentheses 27 close parentheses plus 5 open parentheses 2 x close parentheses to the power of 1 open parentheses 81 close parentheses plus 1 open parentheses 2 x close parentheses to the power of 0 open parentheses 243 close parentheses

space space space space space space space space space space space space space space space space space space equal 32 x to the power of 5 plus 15 cross times 16 open parentheses x close parentheses to the power of 4 plus 90 cross times 8 open parentheses x close parentheses to the power of 3 plus 270 cross times 4 open parentheses x close parentheses to the power of 2 plus 405 open parentheses 2 x close parentheses plus 1 cross times 1 open parentheses 243 close parentheses

space space space space space space space space space space space space space space space space space space equal 32 x to the power of 5 plus 240 x to the power of 4 plus 720 x to the power of 3 plus 4320 x to the power of 2 plus 810 x plus 243(Apply bionomial theorem)

 

answered May 30, 2013 by diane Scholar
0 votes

2.

(x + 15) / (x + 6) ≥ 2

(x + 6 + 9) / (x + 6) ≥ 2

(x + 6) / (x + 6) + 9 / (x + 6) ≥ 2

1 + 9 / (x + 6)≥ 2

Subtract 1 from each side

9 / (x + 6) ≥ 2 -1

9 / (x + 6) ≥ 1

Multiply each side by (x + 6)

9 ≥ (x + 6)

x + 6 ≤ 9

Subtract 6 from each side

x + 6 - 6 ≤ 9 - 6

x ≤ 3.

 

answered May 30, 2013 by diane Scholar
0 votes

The inequality is image

  • Step-1

State the exclude values,These are the values for which denominator is zero.

The exclude value of the inequality is -6.

  • Step - 2

Solve the related equationimage

image

image

image

image

image

Solution of related equation x   = 3.

  • Step - 3

Draw the vertical lines at the exclude value and at the solution to separate the number line into intervals.

  • Step - 4

Now test  sample values in each interval to determine whether values in the interval satisify the inequality.

Test x = -7 in (-∞, -6)

image

image

Above statement is false.

Test x = -1 in (-6, 3)

image

image

Above statement is true.

Test x = 4 in (3, ∞)

image

image

Above statement is false.

Test x = 3

image

image

Above statement is true.

The inequality is satisfied on the open intervals (-6, 3) .Moreover, because image when x = 3, you can conclude that the solution set consists of all real numbers in the intervals (-6, 3] (Be sure to use a closed interval to indicate that can equal 3.)

Number line graph

Solution -6 < ≤ 3.

answered May 30, 2014 by david Expert

Related questions

asked Nov 12, 2014 in CALCULUS by anonymous
asked Nov 12, 2014 in CALCULUS by anonymous
asked Jul 23, 2014 in PRECALCULUS by anonymous
...