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2 years ago

What is the negative solution to this equation? 4x^2+12x=135

Mathematics

1 answer:

Scrat [10]2 years ago

6 0

Answer:

$-\frac{15}{2}$

Step-by-step explanation:

$\mathrm{Subtract\:}135\mathrm{\:from\:both\:sides}$

$4x^2+12x-135=135-135$

$\mathrm{Simplify}$

$4x^2+12x-135=0$

$x1, 2=−b±√b2−4ac2a For a=4, b=12, c=−135x1, 2=−12±√122−4· 4(−135)2· 4$

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For the binomial expansion of (x + y)^10, the value of k in the term 210x 6y k is a) 6 b) 4 c) 5 d) 7

Sloan [31]

Answer:

a) 6

Step-by-step explanation:

Expanding the polynomial using the formula:

$(x+y)^n=\sum_{k=0}^n \binom{n}{k} x^{n-k} y^k$

Also

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

I think you mean $210x^6y^4$

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

For $k=5$

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

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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3 years ago

What is a possible solution to -7x+4>-3x-6

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-7x + 4 > -3x - 6

+7x +7x

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+6 +6

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--- ---

4 4 = 2.5 > x

So anything under 2.5

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3 years ago

What is the negative solution to this equation? 4x^2+12x=135​

What is the negative solution to this equation? 4x^2+12x=135