If you put the value of 4 as n and multiply it by three you will get twelve which satisfies the inequality. Anything larger than 4 will also satisfy the inequality. I really hope this answer helped tell me in the comments.

3 0

2 years ago

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3(8⋅7−45)+5(6⋅4−1) Just checking to see if it is right

AveGali [126]

Answer: 148

Step-by-step explanation:

6 0

2 years ago

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How much are 10 australian dollars worth in Canadian dollars

cupoosta [38]

Answer: 8.94

Step-by-step explanation:

6 0

2 years ago

Help please!<br><br> | 2x^2 + 5x + 3 | > 0<br> solve the inequality

vovikov84 [41]

Factorize the quadratic.

$2x^2 + 5x + 3 = (2x + 3) (x + 1)$

We have $|ab| = |a||b|$ , so

$|2x^2 + 5x + 3| = |2x + 3| |x + 1| > 0$

Now, both $|2x+3|\ge0$ and $|x+1|\ge0$ (since the absolute value of any number cannot be negative), so we just need to worry about when the left side is exactly zero. This happens for

$(2x + 3) (x + 1) = 0 \implies 2x+3 = 0\text{ or }x+1 = 0 \\\\ \implies x = -\dfrac32 \text{ or } x = -1$

So the solution to the inequality is the set

$\left\{x \in \Bbb R \mid x\neq-\dfrac32 \text{ and } x\neq-1\right\}$

3 0

1 year ago