Three consecutive even integers have a sum of -48. Find the largest integer.

Mathematics

1 answer:

kobusy [5.1K]3 years ago

6 0

Answer:

Step-by-step explanation:

{x - some integer}

2x - the first even integer

2x+2 - the second consecutive even integer

2x+4 - the third consecutive even integer (the largest)

2x + 2x+2 + 2x+4 - the sum of three consecutive even integers

2x + 2x+2 + 2x+4 = -48

6x + 6 = -48

÷3 ÷3

2x + 2 = -16

+2 +2

2x+4 = -14

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7 0

3 years ago

Read 2 more answers

Please help me with 2b ASAP. <br> Really appreciate it!!

Bogdan [553]

$f(x)=\dfrac{x^2}{x^2+k^2}$

By definition of the derivative,

$f'(x)=\displaystyle\lim_{h\to0}\frac{\frac{(x+h)^2}{(x+h)^2+k^2}-\frac{x^2}{x^2+k^2}}h$

$f'(x)=\displaystyle\lim_{h\to0}\frac{(x+h)^2(x^2+k^2)-x^2((x+h)^2+k^2)}{h(x^2+k^2)((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{(x+h)^2-x^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2xh+h^2}{h((x+h)^2+k^2)}$

$f'(x)=\dfrac{k^2}{x^2+k^2}\displaystyle\lim_{h\to0}\frac{2x+h}{(x+h)^2+k^2}$

$f'(x)=\dfrac{2xk^2}{(x^2+k^2)^2}$

$\dfrac{k^2}{(x^2+k^2)^2}$ is positive for all values of $x$ and $k$ . As pointed out, $x\ge0$ , so $f'(x)\ge0$ for all $x\ge0$ . This means the proportion of occupied binding sites is an increasing function of the concentration of oxygen, meaning the presence of more oxygen is consistent with greater availability of binding sites. (The question says as much in the second sentence.)