Help me find the perimeter of this square. if you

Ответ:

123abc321cbaabc

15.07.2021

Mathematics

396

Step-by-step explanation:

It's a square.

That means that all four sides are equal.

So the two expressions you have been given are equal.

2.5x + 76.5 = 12x - 9 Subtract 2.5x from both sides.

-2.5x -2.5x

76.5 = 9.5x - 9 Add 9 to both sides

9 9

85.5 = 9.5x Divide by 9.5

85.5/9.5 = x

x = 9

That is just the value for x. It is not the answer

Side = 12x - 9

Side = 12*9 - 9

Side = 108 - 9

Side = 99

The perimeter = 4 * Side

The perimeter = 4 * 99

Perimeter = 396

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Ответ:

michelle7511

10.09.2020

Mathematics

Answer

$\left( -\dfrac{2 \sqrt{3} }{3},\ \dfrac{2 \sqrt{3} }{3}\right)$

explanation

to find the largest open interval where f is in decreasing, we find the first derivative, f'. then we find the critical numbers, aka where f'(x) = 0 or undefined. after that, we determine where f' is negative and positive using the critical numbers as region boundaries.so we first find the first derivative.

differentiate f using product rule:

$\begin{aligned} f(x) & = x^3 - 4x \\ f'(x) & = 3x^2 - 4 \end{aligned}$

find where f'(x) = 0; set f'(x) = 0 and solve for x

$\begin{aligned} 3x^2 - 4 & = 0\\ 3x^2 & = 4 \\ x^2 & = \tfrac{4}{3} \\ x & = \pm \sqrt{ \tfrac{4}{3} } \\ x & = \pm \tfrac{2}{\sqrt{3}} \\ x & = \pm \tfrac{2\sqrt{3}}{3} & & \text{\footnotesize multiply numerator and denom. by $\sqrt{3}$ to rationalize} \end{aligned}$

those two numbers are critical numbers. set up a sign chart and test the three intervals that form from these two critical numbers. these three intervals are

$\left(-\infty, -\tfrac{2\sqrt{3}}{3} \right),\\ \left(-\tfrac{2\sqrt{3}}{3}, \tfrac{2\sqrt{3}}{3}\right), \\ \left(\tfrac{2\sqrt{3}}{3}, \infty \right)$

if f' is negative on an interval i ⇒ f is decreasing on interval iif f' is positive on an interval i ⇒ f is increasing on interval i

evaluate f' at numbers in each interval to test the sign of f'. for example, in the interval $\left(-\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3} \right)$ , we can test f'(0) (since 0 is in that open interval) and that it is negative.

$\begin{array}{c | c | c} \text{interval} & \text{sign of $f'$} & \text{$f$ inc/dec/neither}\\ \left(-\infty, -\tfrac{2\sqrt{3}}{3} \right) & (+) & \text{increasing} \\ \left(-\tfrac{2\sqrt{3}}{3}, \tfrac{2\sqrt{3}}{3} \right) & & \text{decreasing} \\ \left(\tfrac{2\sqrt{3}}{3}, \infty \right) & (+) & \text{increasing}\end{array}$

the only interval on which f is decreasing is

$\left(-\tfrac{2\sqrt{3}}{3}, \tfrac{2\sqrt{3}}{3} \right)$

and, therefore, that is the answer.

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