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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

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