Math, please help! Will give brainliest.

An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu

exis [7]

Take the derivative with respect to t
$- w \sin(wt) + \sqrt{8} w cos(wt)$
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
$0 = -w \sin(wt) + \sqrt{8} w cos(wt)$
divide by w
$0 =- \sin(wt) + \sqrt{8} cos(wt)$
we add sin(wt) to both sides

$\sin(wt)= \sqrt{8} cos(wt)$
divide both sides by cos(wt)
$\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}$ OR $\\$ $wt=arctan( { \frac{1}{ \sqrt{2} } )$
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

$I=cos(2n \pi -2arctan( \sqrt{2} ))$
since 2npi is just the period of cos
$cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 
$
substituting our second soultion we get
$I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))$
since 2npi is the period
$I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}$
so the maximum value = $\frac{1}{3}$
minimum value = $- \frac{1}{3}$

4 0

2 years ago

Acellus. Solve for x. x = ?

Makovka662 [10]

Answer:

according to basic proportionality theorem

$\qquad\sf {:}\longrightarrow 25/35=x/63$

$\qquad\sf {:}\longrightarrow 5/7=x/63$

$\qquad\sf {:}\longrightarrow 7x=63×5=315$

$\qquad\sf {:}\longrightarrow x=315/7=45$

5 0

2 years ago

a chord of 12cm long of 8cm is away from the of the circle what is the lenght of the chord which is 6cm away from the centre

lana66690 [7]

Answer:

The length of the chord is 16 cm

Step-by-step explanation:

Mathematically, a line from the center of the circle to a chord divides the chord into 2 equal portions

From the first part of the question, we can get the radius of the circle

The radius form the hypotenuse, the two-portions of the chord (12/2 = 6 cm) and the distance from the center to the chord forms the other side of the triangle

Thus, by Pythagoras’ theorem; the square of the hypotenuse equals the sum of the squares of the two other sides

Thus,

r^2 = 8^2 + 6^2

r^2= 64 + 36

r^2 = 100

r = 10 cm

Now, we want to get a chord length which is 6 cm away from the circle center

let the half-portion that forms the right triangle be c

Using Pythagoras’ theorem;

10^2 = 6^2 + c^2

c^2 = 100-36

c^2 = 64