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Strike441 [17]
3 years ago
7

( Don't Report Pleaseee) Yoooo do you guise agree me on that japan should make a skateboarding anime show like bro like i feel l

ike that would be good. lol maybe it is just me
Mathematics
2 answers:
viktelen [127]3 years ago
8 0

Answer:

yea for real they should ngl I love anime.

Stels [109]3 years ago
8 0

Answer:

Maybe they should but they would have to have some good characters

You might be interested in
Of the 22 students in the class, 16 of them are boys.
Eddi Din [679]
22 students...16 boys

probability student will be a boy is 16/22 which reduces to 8/11 or 72.7%
6 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 <em>x: </em>

\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 <em>x</em> = 0 and <em>x</em> = 2.

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

To solve for <em>x</em>, 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 <em>x: </em>

\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 <em>a</em> and <em>b</em> > 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 <em>a</em> = 2 and <em>b</em> = 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 <em>x</em> = 0.5 and <em>x</em> = 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
An amoeba is 0.8 millimeter in length. At the science museum there is a scale model of the amoeba that is 160 millimeters in len
lisabon 2012 [21]
The scale factor is 200
4 0
3 years ago
I am not understanding the question. ​
artcher [175]

Answer:

This question is asking you to find the ending position of a shape (particle) that is moving horizontally along the horizontal axis by finding its velocity with the equation shown. When t=0, the shape is 4 units to the left of the vertical axis, so it wants you to find where on the horizontal axis the shape is when t = 8. Hope this helped!!!

6 0
3 years ago
Read 2 more answers
The length of a rectangular football field, including both end zones, is 120 yards. The area of the field is 57,600 square feet.
egoroff_w [7]

Answer:

3a.False,the width is 53.33 yards

3b.true, the Length is 120 yards=360 feet

3c.True, width of the field is 160 feet.

3d. True,rea of the field is 6,400 square yards.

Step-by-step explanation:

Hello , I think I can help you with this

to solve this you must know the equivalence between feet and yard, which is

3 feet = 1 yard

Also, the area of  a rectangle is the product of its length by its width

Area=length *width

Step 1

convert 120 yards into feet using a rule of three

3 feet = 1 yard

x feet= 120 yards

3/1=x/120

x=(120*3)/1

x=360 feet

so

120 yards= 360 feet

Step 2

isolate the width from the area´s equation to find out it.

Area=57600 square feet

length=360 feet

Area=length *width

width=Area/length

Width=57600 square feet/360 feet

Width=160 feet

Step 3

convert 160 feet into yards

1 yard= 3 feet

x yard= 160 feet

1/3=x/160

x=(1*160)/3

x=53.33 yards

width=53.33 yards

the area is (square yards)

Area=120 yards*53.33 yards

Area=6400 square yards

Step 4

Answer the questions.

3a.False,the width is 53.33 yards

3b.true, the Length is 120 yards=360 feet

3c.True, width of the field is 160 feet.

3d. True,rea of the field is 6,400 square yards.

Have a nice day.

7 0
3 years ago
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