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3 years ago

Use a calculator to solve for theta in the given interval

Mathematics

1 answer:

Allushta [10]3 years ago

4 0

Answer:

d. $1.82\: or\:4.46$

Step-by-step explanation:

The given trigonometric equation is;

$\sec \theta=-4.0545$

Recall that;

$\cos \theta=\frac{1}{\sec \theta}$

$\implies \cos \theta=\frac{1}{-4.0545}$

$\implies \cos \theta=-0.2466$

The cosine function is negative in the second and third quadrant.

$\theta=\pi-\cos^{-1}(0.2466)\: or\:\pi+\cos^{-1}(0.2466)$

$\theta=1.82\: or\:4.46$

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Consider the function f(x, y) = x2 + xy + y2 defined on the unit disc, namely, d = {(x, y)| x2 + y2 ≤ 1}. use the method of lagr

Pani-rosa [81]

First we note that the partial derivatives vanish simultaneously at one point:

$\begin{cases}f_x=2x+y=0\\f_y=x+2y=0\end{cases}\implies(x,y)=(0,0)$

so there is one critical point within the region $D$ .

The Lagrangian is

$L(x,y,\lambda)=x^2+xy+y^2+\lambda(x^2+y^2-1)$

and has partial derivatives

$L_x=2x+y+2\lambda x$
$L_y=x+2y+2\lambda y$
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Set each partial derivative to 0 to find the possible critical points within the disk $D$ . Then we notice that

$yL_x=2xy+y^2+2\lambda xy=0$
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$\implies xL_y-yL_x=x^2-y^2=0$

Since $x^2+y^2=1$ , we have

$x^2-y^2=x^2+y^2-2y^2=0\implies1=2y^2\implies y^2=\dfrac12\implies y=\pm\dfrac1{\sqrt2}$

And since $x^2-y^2=0$ , or $x^2=y^2$ , we also have

$x=\pm\dfrac1{\sqrt2}$

So we have four possible additional critical points to consider:

$f(0,0)=0$
$f\left(-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac32$
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$f\left(\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)=\dfrac12$
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8 0

3 years ago

Find the area of a rectangle with a width of x3 + 5x - 9 and a length of x2 + 8.

Drupady [299]

Answer:

A = x⁵ + 13x³ - 9x² + 40x - 72

Step-by-step explanation:

The formula for the area of a rectangle is length X width, or A = lw.

Substitute the equations for the sides into the formula:

A = lw

A = (x² + 8)(x³ + 5x - 9) <= use the distributive property

A = x⁵ + 5x³ -9x² + 8x³ + 40x - 72 <= collect like terms