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sukhopar [10]
2 years ago
12

5x−4≥12 OR 12x+5≤−4 please answer asap and right and explain like im 5 how to solve OR problems ill give brainliest and a lot of

points

Mathematics
1 answer:
weeeeeb [17]2 years ago
4 0

Answer: x>_3.2 OR x<_ -0.75

Step-by-step explanation: first break down your compound inequality. 5x-4>_12

You first cancel out your constants by adding 4 to both sides. Now you’re left with 5x>_16 then to cancel five you have to divide on both sides by five which equals to 3.2. Then, x>_ 3.2.

Next you do your second part, 12x+5<_-4

So first cancel out the constant of 5 by subtracting 5 on both sides, making the equation 12x<_-9. Now, you divide by 12 on both sides, making it -9/12. Which effectively is -0.75. Therefor, the answer being x<_ -0.75. Add the two together x>_3.2 OR x<_0.75

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I think you mean 210x^6y^4

We can deduce that this term will be located somewhere in the middle. So I will calculate k= 5; k=6 \text{ and } k =7.

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$\binom{10}{5} (y)^{10-5} (x)^{5}=\frac{10!}{(10-5)! 5!}(y)^{5} (x)^{5}= \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5! }{5! \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 } \\ =\frac{30240}{120} =252 x^{5} y^{5}$

Note that we actually don't need to do all this process. There's no necessity to calculate the binomial, just x^{n-k} y^k

For k=6

$\binom{10}{6} \left(y\right)^{10-6} \left(x\right)^{6}=\frac{10!}{(10-6)! 6!}\left(y\right)^{4} \left(x\right)^{6}=210 x^{6} y^{4}$

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