67.1951 round to the nearest hundredth

The graph of the function g(x) is shown on the coordinate plane below. For what value of x does g(x)=0

Murrr4er [49]

The answer is -2.5 there you go

8 0

3 years ago

Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1

mafiozo [28]

Answer:

(a). y'(1)=0 and y'(2) = 3

(b). $y'(t)=kb2t\cos(bt^2)$

(c). $b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value $x_{0}$ which lies in the domain of f where the derivative is 0.

Therefore, y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function, $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

$y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]$

Applying chain rule,

$y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)$

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0 (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

$2k\pi(1) = 3$$

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\ k = \frac{3}{2\pi}$

7 0

4 years ago

Is anybody else’s brainly app messing up recently? Every time I try to look a question up, it says something went wrong..

Tanya [424]

Answer:

Same thing here.

4 0

3 years ago

Answer this maths watch question step-by-step

Scilla [17]

The value of <u>a</u> is 2 and the value of <u>b</u> is 6.

<h3>Factoring</h3>

In math, factoring or factorization is used to write an algebraic expression in factors. There are some rules for factorization. One of them is a factor out a common term for example: x²-x= x(x-1), where x is a common term.

You should factor the given expression.

$2-\frac{x+3}{x+1}-\frac{x-5}{x-1}\\ \\ 2+\frac{-\left(x+3\right)\left(x-1\right)-\left(x-5\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ \\ \mathrm{Apply\:the\:fraction\:rule}:\quad \:a+\frac{b}{c}=\frac{ac+b}{c}\\ \\ \frac{2\left(x+1\right)\left(x-1\right)-\left(x+3\right)\left(x-1\right)-\left(x-5\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ \\ \frac{2x+6}{\left(x+1\right)\left(x-1\right)}=\frac{2x+6}{x^2-1}$