Can someone answer this? It’s 9th grade algebra

Mathematics

1 answer:

Colt1911 [192]3 years ago

5 0

Answer:

Step-by-step explanation:

We want to solve the inequality:

$\displaystyle 5.1(3 + 2.2x) > -14.25 - 6(1.7x+4)$

Distribute:

$(15.3+11.22x) > -14.25 + (-10.2x-24)$

Combine like terms:

$15.3 + 11.22x > -10.2x-38.25$

We can add 10.2<em>x</em> to both sides:

$15.3 + 21.42x > -38.25$

Subtract 15.3 from both sides:

$21.42x > -53.55$

And divide both sides by 21.42. Hence:

$\displaystyle x > -2.5$

This means all values greater than -2.5. Hence, in interval notation:

$(-2.5, \infty)$

In conclusion, our answer is C.

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Answer:

$\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }$

Step-by-step explanation:

Hello, please find below my work.

$2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0$

$\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}$