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3 years ago

14

If h is a function and -7 is an element in its domain, then which of the following statements correctly describes the expression

h(-7)?

2 answers:

Triss [41]3 years ago

8 0

-7.675628 times 86782.88

krok68 [10]3 years ago

7 0

Answer:

The first choice: The expression h(-7) represents the output of h corresponding to the input of -7.

Step-by-step explanation:

In function f(5), f(5) means the value of the function at x = 5.

Answer:

The first choice: The expression h(-7) represents the output of h corresponding to the input of -7.

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Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

5 0

3 years ago

3. What is the slope of the line passing through (15,-5) and (3, -8)?

Andrej [43]

M= -8 - -5/ 3 - 15= -8+5/3-15= -3/-12 = 3/12 = 1/4

The slope is 1/4.
The slope formula is m=y2-y1/x2-x1. I plugged in -8=y2, -5=y1, 3=x2, and 15=x1. Then I solved it.

6 0

3 years ago

Read 2 more answers

What is the measure of the angle ?

Roman55 [17]

Answer:

Answer → C) 121°

Step-by-step explanation:

» Let the wanted angle be x

${ \tt{x = 121 \degree}} \\ { \tt{ \{alternate \: angles \}}}$

4 0

2 years ago

Mr Smith's art class took a bus trip to an art museum. The bus averaged 65 miles per hour on the highway and 25 miles per hour i

Leya [2.2K]

Let x be the distance traveled on the highway and y the distance traveled in the city, so:
$\left \{ {{x+y=375} \atop { \frac{1}{65}x+ \frac{1}{25}y =7}} \right.$

Now, the system of equations in matrix form will be:
$\left[\begin{array}{ccc}1&1&\\ \frac{1}{65} & \frac{1}{25} &\end{array}\right] \left[\begin{array}{ccc}x&\\y&\end{array}\right] = \left[\begin{array}{ccc}375&\\7&\end{array}\right]$

Next, we are going to find the determinant:
$D= \left[\begin{array}{ccc}1&1\\ \frac{1}{65} & \frac{1}{25} \end{array}\right] =(1)( \frac{1}{25}) - (1)( \frac{1}{65} )= \frac{8}{325}$
Next, we are going to find the determinant of x:
$D_{x} = \left[\begin{array}{ccc}375&1\\7& \frac{1}{25} \end{array}\right] = (375)( \frac{1}{25} )-(1)(7)=8$

Now, we can find x:
$x= \frac{ D_{x} }{D} = \frac{8}{ \frac{8}{325} } =325mi$

Now that we know the value of x, we can find y:
$y=375-325=50mi$

Remember that time equals distance over velocity; therefore, the time on the highway will be:
$t_{h} = \frac{325}{65} =5hours$
An the time on the city will be:
$t_{c} = \frac{50}{25} =2hours$

We can conclude that the bus was five hours on the highway and two hours in the city.

8 0

3 years ago

The function f is continuous on the closed interval [1,15] and has the values shown on the table above. Let g(x) = ∫f(t) dt [1,x

Anna11 [10]

We're looking for the two values being subtracted here. One of these values is easy to find:

<span>g(1) = ∫f(t)dt = 0</span><span>
since taking the integral over an interval of length 0 is 0.

The other value we find by taking a Left Riemann Sum, which means that we divide the interval [1,15] into the intervals listed above and find the area of rectangles over those regions:

</span><span>Each integral breaks down like so:

(3-1)*f(1)=4

(6-3)*f(3)=9

(10-6)*f(6)=16

(15-10)*f(10)=10.

So, the sum of all these integrals is 39, which means g(15)=39.

Then, g(15)-g(1)=39-0=39.
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!

3 0

3 years ago

Read 2 more answers