All categories

3 years ago

PLZ HELP!!! A Circle has a radius of 6x9y5 Cm.

Mathematics

1 answer:

VMariaS [17]3 years ago

4 0

Hope this helps you with your question

You might be interested in

Help me plz!!!!!!!!!!!

svetlana [45]

The first one is the answer

7 0

3 years ago

Read 2 more answers

Find the slope of the line that goes through the following pair of points.

Juli2301 [7.4K]

It’s a bro trust me on this one

7 0

3 years ago

If a and b are positive numbers, find the maximum value of f(x) = x^a(2 − x)^b on the interval 0 ≤ x ≤ 2.

Ad libitum [116K]

Answer:

The maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Step-by-step explanation:

Answer:

Step-by-step explanation:

We are given the function:

$\displaystyle f(x) = x^a (2-x)^b \text{ where } a, b >0$

And we want to find the maximum value of f(x) on the interval [0, 2].

First, let's evaluate the endpoints of the interval:

$\displaystyle f(0) = (0)^a(2-(0))^b = 0$

And:

$\displaystyle f(2) = (2)^a(2-(2))^b = 0$

Recall that extrema occurs at a function's critical points. The critical points of a function at the points where its derivative is either zero or undefined. Thus, find the derivative of the function:

$\displaystyle f'(x) = \frac{d}{dx} \left[ x^a\left(2-x\right)^b\right]$

By the Product Rule:

$\displaystyle \begin{aligned} f'(x) &= \frac{d}{dx}\left[x^a\right] (2-x)^b + x^a\frac{d}{dx}\left[(2-x)^b\right]\\ \\ &=\left(ax^{a-1}\right)\left(2-x\right)^b + x^a\left(b(2-x)^{b-1}\cdot -1\right) \\ \\ &= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right] \end{aligned}$

Set the derivative equal to zero and solve for x:

$\displaystyle 0= x^a\left(2-x\right)^b \left[\frac{a}{x} - \frac{b}{2-x}\right]$

By the Zero Product Property:

$\displaystyle x^a (2-x)^b = 0\text{ or } \frac{a}{x} - \frac{b}{2-x} = 0$

The solutions to the first equation are x = 0 and x = 2.

First, for the second equation, note that it is undefined when x = 0 and x = 2.

To solve for x, we can multiply both sides by the denominators.

$\displaystyle\left( \frac{a}{x} - \frac{b}{2-x} \right)\left((x(2-x)\right) = 0(x(2-x))$

Simplify:

$\displaystyle a(2-x) - b(x) = 0$

And solve for x:

$\displaystyle \begin{aligned} 2a-ax-bx &= 0 \\ 2a &= ax+bx \\ 2a&= x(a+b) \\ \frac{2a}{a+b} &= x \end{aligned}$

So, our critical points are:

$\displaystyle x = 0 , 2 , \text{ and } \frac{2a}{a+b}$

We already know that f(0) = f(2) = 0.

For the third point, we can see that:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(2- \frac{2a}{a+b}\right)^b$

This can be simplified to:

$\displaystyle f\left(\frac{2a}{a+b}\right) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

Since a and b > 0, both factors must be positive. Thus, f(2a / (a + b)) > 0. So, this must be the maximum value.

To confirm that this is indeed a maximum, we can select values to test. Let a = 2 and b = 3. Then:

$\displaystyle f'(x) = x^2(2-x)^3\left(\frac{2}{x} - \frac{3}{2-x}\right)$

The critical point will be at:

$\displaystyle x= \frac{2(2)}{(2)+(3)} = \frac{4}{5}=0.8$

Testing x = 0.5 and x = 1 yields that:

$\displaystyle f'(0.5) >0\text{ and } f'(1)$

Since the derivative is positive and then negative, we can conclude that the point is indeed a maximum.

Therefore, the maximum value of f(x) occurs at:

$\displaystyle x = \frac{2a}{a+b}$

And is given by:

$\displaystyle f_{\text{max}}(x) = \left(\frac{2a}{a+b}\right)^a\left(\frac{2b}{a+b}\right)^b$

5 0

3 years ago

A truck left a station at 6:00am traveling at a rate of 45km/h. A van left the same place 2 hours later heading in the same dire

OverLord2011 [107]

The Van would catch up with the truck at 6:00 pm

Speed is the ratio of distance travelled to time taken. It is given by:

Speed = distance / time

Let us say the van catch up with the truck after t hours at a distance d.

For the truck:

45 = d/t

d = 45t

For the Van:

54 = d / (t - 2)

d = 54t - 108

45t = 54t - 108

9t = 108

t = 12 hours

The Van would catch up with the truck at 6:00 pm

Find out more on speed at: brainly.com/question/3004254

6 0

3 years ago

Read 2 more answers

Marco and Drew stacked boxes on a shelf. Marco lifted 9 boxes and Drew lifted 14 boxes. The boxes that Drew lifted each weighed

MrRa [10]

The correct answer is b. 14(m-8).

This is because it says that the weight of the boxes drew lifted were 8 pounds less than what marco lifted. This can be represented by m-8. You then need to multiply this by the amount of boxes that drew lifted, or 14.

8 0

3 years ago