Solve the following absolute value inequalities and partial fractions: |5x+4| ext 2, and…

Question

Solve the following absolute value inequalities and partial fractions: |5x+4| ext<= 1, |5x−7| > 2, and resolve rac6x−9(x−3)(x−2) into partial fractions.

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the problems:

49. Solve $|5x+4| \le 1$ .

Rewrite the absolute value inequality.
$-1 \le 5x+4 \le 1$
Subtract 4 from all parts of the inequality.
$-1 - 4 \le 5x \le 1 - 4$ $-5 \le 5x \le -3$
Divide all parts by 5.
$\frac{-5}{5} \le x \le \frac{-3}{5}$ $-1 \le x \le -\frac{3}{5}$
Express the solution in interval notation.
$x \in \left[-1, -\frac{3}{5}\right]$
The correct option is B.

50. Solve $|5x-7| > 2$ .
Rewrite the absolute value inequality as two separate inequalities.
$5x-7 > 2 \quad \text{or} \quad 5x-7 < -2$
Solve the first inequality.
$5x-7 > 2$ $5x > 2+7$ $5x > 9$ $x > \frac{9}{5}$
Solve the second inequality.
$5x-7 < -2$ $5x < -2+7$ $5x < 5$ $x < \frac{5}{5}$ $x < 1$
Combine the solutions and express in interval notation.
$x \in (-\infty, 1) \cup \left(\frac{9}{5}, \infty\right)$
The correct option is B.

51. Resolve $\frac{6x-9}{(x-3)(x-2)}$ into partial fractions.
Set up the partial fraction decomposition.
$\frac{6x-9}{(x-3)(x-2)} = \frac{A}{x-3} + \frac{B}{x-2}$
Multiply both sides by $(x-3)(x-2)$ .
$6x-9 = A(x-2) + B(x-3)$
Solve for A and B.
Let $x=2$ :
$6(2)-9 = A(2-2) + B(2-3)$ $12-9 = 0 + B(-1)$ $3 = -B \implies B = -3$
Let $x=3$ :
$6(3)-9 = A(3-2) + B(3-3)$ $18-9 = A(1) + 0$ $9 = A$
Substitute A and B back into the decomposition.
$\frac{6x-9}{(x-3)(x-2)} = \frac{9}{x-3} + \frac{-3}{x-2}$ $\frac{6x-9}{(x-3)(x-2)} = \frac{9}{x-3} - \frac{3}{x-2}$
The correct option is B.

52. Resolve $\frac{8x+4}{x^2-4}$ into partial fractions.
Factor the denominator.
$x^2-4 = (x-2)(x+2)$
Set up the partial fraction decomposition.
$\frac{8x+4}{(x-2)(x+2)} = \frac{A}{x-2} + \frac{B}{x+2}$
Multiply both sides by $(x-2)(x+2)$ .
$8x+4 = A(x+2) + B(x-2)$
Solve for A and B.
Let $x=2$ :
$8(2)+4 = A(2+2) + B(2-2)$ $16+4 = A(4) + 0$ $20 = 4A \implies A = 5$
Let $x=-2$ :
$8(-2)+4 = A(-2+2) + B(-2-2)$ $-16+4 = 0 + B(-4)$ $-12 = -4B \implies B = 3$
Substitute A and B back into the decomposition.
$\frac{8x+4}{x^2-4} = \frac{5}{x-2} + \frac{3}{x+2}$
The correct option is B.

53. Find the general term $T_{r+1}$ of the binomial expansion of $\left(2x + \frac{1}{x}\right)^8$ .
Use the binomial theorem formula for the $(r+1)$ -th term
$T_{r+1} = \binom{n}{r} a^{n-r} b^r$ .
Here, $n=8$ , $a=2x$ , and $b=\frac{1}{x}$ .
Substitute the values into the formula.
$T_{r+1} = \binom{8}{r} (2x)^{8-r} \left(\frac{1}{x}\right)^r$
Simplify the expression.
$T_{r+1} = \binom{8}{r} 2^{8-r} x^{8-r} x^{-r}$ $T_{r+1} = \binom{8}{r} 2^{8-r} x^{8-r-r}$ $T_{r+1} = \binom{8}{r} 2^{8-r} x^{8-2r}$
The correct option is B.

Solve the following absolute value inequalities and partial fractions: |5x+4| ext<= 1, |5x-7| > 2, and resolve rac6x-9(x-3)(x-2) into partial fractions.

Answer

Step-by-step solution

Rewrite the absolute value inequality.

Subtract 4 from all parts of the inequality.

Divide all parts by 5.

Express the solution in interval notation.

Rewrite the absolute value inequality as two separate inequalities.

Solve the first inequality.

Solve the second inequality.

Combine the solutions and express in interval notation.

Set up the partial fraction decomposition.

Multiply both sides by $(x-3)(x-2)$ .

Solve for A and B.

Substitute A and B back into the decomposition.

Factor the denominator.

Set up the partial fraction decomposition.

Multiply both sides by $(x-2)(x+2)$ .

Solve for A and B.

Substitute A and B back into the decomposition.

Use the binomial theorem formula for the $(r+1)$ -th term

Substitute the values into the formula.

Simplify the expression.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Solve the following absolute value inequalities and partial fractions: |5x+4| ext<= 1, |5x-7| > 2, and resolve rac6x-9(x-3)(x-2) into partial fractions.

Answer

Step-by-step solution

Rewrite the absolute value inequality.

Subtract 4 from all parts of the inequality.

Divide all parts by 5.

Express the solution in interval notation.

Rewrite the absolute value inequality as two separate inequalities.

Solve the first inequality.

Solve the second inequality.

Combine the solutions and express in interval notation.

Set up the partial fraction decomposition.

Multiply both sides by (x−3)(x−2)(x-3)(x-2)(x−3)(x−2).

Solve for A and B.

Substitute A and B back into the decomposition.

Factor the denominator.

Set up the partial fraction decomposition.

Multiply both sides by (x−2)(x+2)(x-2)(x+2)(x−2)(x+2).

Solve for A and B.

Substitute A and B back into the decomposition.

Use the binomial theorem formula for the (r+1)(r+1)(r+1)-th term

Substitute the values into the formula.

Simplify the expression.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Multiply both sides by $(x-3)(x-2)$ .

Multiply both sides by $(x-2)(x+2)$ .

Use the binomial theorem formula for the $(r+1)$ -th term