What is the domain of the function..... {TIMED PLS HURRY}

Resolving inequalities 6/5x-12<2-3/2x

Anna007 [38]

Answer: x < $\frac{140}{27}$

Step-by-step explanation:

$\frac{6}{5}x - 12 < 2 - \frac{3}{2}x$

$10(\frac{6}{5}x - 12) < 10(2 - \frac{3}{2}x)$

2(6)x - 10(12) < 10(2) - 5(3)x

12x - 120 < 20 - 15x

+15x +15x

27x - 120 < 20

+120 +120

27x < 140

$\frac{27x}{27} < \frac{140}{27}$

x < $\frac{140}{27}$

3 0

4 years ago

Graph the function y=-|x|+9

MakcuM [25]

Answer:

Graph is attached :)

Step-by-step explanation:

Hope that helps!

8 0

2 years ago

Find the solution set to each inequality. Express the solution in set notation.

daser333 [38]

The solution to the inequality 6m + 2 > -27 is m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is p > 8

The given inequality is:

6m + 2 > - 27

Subtract 2 to both sides of the inequality

6m + 2 - 2 > -27 - 2

6m > -29

Divide both sides by 6

$\frac{6m}{6}> \frac{-29}{6} \\m > \frac{-29}{6}\\m > -4.83$

For the inequality 8(p-6)>4(p-4)

Expand the inequality using the distributive rule

8p - 48 > 4p - 16

Collect like terms

8p - 4p > -16 + 48

4p > 32

Divide both sides of the inequality 4

$\frac{4p}{4} > \frac{32}{4}\\p > 8$

The solution to the inequality 6m + 2 > -27 is m > -4.33

The solution to the inequality 8(p-6)>4(p-4) is p > 8

Learn more here: brainly.com/question/15816805

3 0

3 years ago

I need help in partial fraction!! With simple explaination would be nice!

WITCHER [35]

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}$

The degree of the numerator (4) is larger than the degree of the denominator (3), so first you need to divide. (Added screenshot of long division procedure.)

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x-\dfrac{4x^2-17x+10}{x(x^2-3)}$

Now the second term can be decomposed into partial fractions.

$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1}x+\dfrac{r_2x+r_3}{x^2-3}$
$\dfrac{4x^2-17x+10}{x(x^2-3)}=\dfrac{r_1(x^2-3)+x(r_2x+r_3)}{x(x^2-3)}$
$4x^2-17x+10=r_1(x^2-3)+x(r_2x+r_3)$
$4x^2-17x+10=(r_1+r_2)x^2+r_3x-3r_1$
$\implies\begin{cases}r_1+r_2=4\\r_3=-17\\-3r_1=10\end{cases}\implies r_1=-\dfrac{10}3,r_2=\dfrac{22}3,r_3=-17=-\dfrac{51}3$
$\implies\dfrac{4x^2-17x+10}{x(x^2-3)}=-\dfrac{10}{3x}+\dfrac{22x-51}{x^2-3}$

So

$\dfrac{x^4-7x^2+17x-10}{x(x^2-3)}=x+\dfrac{10}{3x}-\dfrac{22x-51}{x^2-3}$