PLEASE HELP ASAP (: Fill in the blank to complete this trig ratio. sin a =

Which of the following numbers is not part of the solution to x ≤ -2?

podryga [215]

Answer:

-1.76

Step-by-step explanation:

Since it is in the negatives, the higher the number it is, such as -8, will be less than -2

4 0

2 years ago

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Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

2 years ago

Bradley answered 80% of the questions on his science test correctly. There were 30 questions on the test, and all the questions

son4ous [18]

Answer:

24 correct questions

Step-by-step explanation:

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2 years ago

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Type the correct answer in the box. Use numerals instead of words. If necessary, use / for the fraction bar. The table shows a p

topjm [15]

What is the question ????????????

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2 years ago

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Help with trigonometry

Oksanka [162]

Answer:

○B. $\frac{\sqrt{6}}{2}$

Step-by-step explanation:

Trigonometric Identities

$\frac{y}{r} = sin\:θ$

$\frac{x}{r} = cos\:θ$

$\frac{y}{x} = tan\:θ$

$\frac{r}{x} = sec\:θ$

$\frac{r}{y} = csc\:θ$

$\frac{x}{y} = cot\:θ$

Radius Formula

${y}^{2} + {x}^{2} = {r}^{2}$

${[-\sqrt{15}]}^{2} + {[-\sqrt{10}]}^{2} = {r}^{2} → 15 + 10 = {r}^{2} → 25 = {r}^{2}\\ \\ 5 = r$

* Since we are talking about radii, we only want the NON-NEGATIVE root.

In this case, we will not be using the radius in our ratio, according to the trigonometric identity above because we are using the tangent ratio:

$\frac{\sqrt{6}}{2} = \frac{-\sqrt{15}}{-\sqrt{10}} = \sqrt{1\frac{1}{2}}$

I am joyous to assist you anytime.

4 0

2 years ago