One half times one third

Let f(x) = xe^-x+ ce^-x, where c is a positive constant. For what positive value of c does f have an absolute

Illusion [34]

Answer:

$c=6$

Step-by-step explanation:

The absolute maximum of a continuous function $f(x)$ is where $f'(x)=0$ . Therefore, we must differentiate the function and then set $x=-5$ and $f'(x)=0$ to determine the value of $c$ :

$f(x)=xe^{-x}+ce^{-x}$

$f'(x)=-xe^{-x}+e^{-x}-ce^{-x}$

$0=-(-5)e^{-(-5)}+e^{-(-5)}-ce^{-(-5)}$

$0=5e^{5}+e^{5}-ce^{5}$

$0=e^5(5+1-c)$

$0=6-c$

$c=6$

Therefore, when $c=6$ , the absolute maximum of the function is $x=-5$ .

I've attached a graph to help you visually see this.

7 0

3 years ago

Please help me solve this

lesantik [10]

Speed = distance over time.
300/5 is the same as 300÷5, which is 60

5 0

4 years ago

-3x - y = 0 and -8x +5y=7

andrezito [222]

Answer:

-23x = 7

Step-by-step explanation:

-3x - y =0

×5. ×5 ×5

-15x -5y = 0

-8x +5y = 7

-15x -5y -0

-23x = 7

8 0

3 years ago

Confused on the function and method for this... help me out please.

mezya [45]

Answer:

∠ YZX ≈ 19.7°

Step-by-step explanation:

using the sine ratio in the right triangle.

sin YZX = $\frac{opposite}{hypotenuse}$ = $\frac{XY}{YZ}$ = $\frac{7.5}{22.3}$ , then

∠ YZX = $sin^{-1}$ ( $\frac{7.5}{22.3}$ ) ≈ 19.7° ( to 3 significant figures )

8 0

2 years ago

Help me please ;-;!!!!! Use a coordinate grid to create a map of a town with at least five different locations, such as a house,

Sholpan [36]

A. Our diagram have six location: my house (2,1), a movie theater (5,3), the school (6,6), the police station (3,-2), the airport (-3,3), and the mall (-2,-3)

B. Distance between my house and the movie theater:
To calculate the distance we are going to use the points (2,1) and (5,3) to create a right triangle with tow legs of measures 3 and 2; the hypotenuse of the right triangle will be the distance between my house and the movie theater.
Using the Pythagorean theorem:
$d^2=3^2+2^2$
$d^2=9+4$
$d^2=13$
$d= \sqrt{13}$
We can conclude that the distance between my house and the movie theater is $\sqrt{13}$

Our second distance is the distance between the police station and the airport:
This time we are using the points (3,-2) and (-3,3) to create a right triangle with legs of measures 6 and 5; the hypotenuse of our triangle will be the the distance between our points:
$d^2=6^2+5^2$
$d^2=36+25$
$d^2=61$
$d= \sqrt{61}$
We can conclude that the distance between the police station and the airport is $\sqrt{61}$

C. Distance form the park to the Fire Station:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance from the park to the Fire Station is $\sqrt{10}$

Distance from the stadium to the mall:
$d^2=2^2+1^2$
$d^2=4+1$
$d^2=5$
$d= \sqrt{5}$
We can conclude that the distance from the stadium to the mall is $\sqrt{5}$

D. What is the distance between your house and the mall?
Answer:
$d^2=4^2+4^2$
$d^2=16+16$
$d^2=32$
$d= \sqrt{32}$
$d=4 \sqrt{2}$
We can conclude that the distance between my house and the mall is $4 \sqrt{2}$

What is the distance between the movie theater and the school?
answer:
$d^2=3^2+1^2$
$d^2=9+1$
$d^2=10$
$d= \sqrt{10}$
We can conclude that the distance between the movie theater and the school is $\sqrt{10}$

7 0

3 years ago