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3 years ago

Whats a square + b square I don't know thats why I am asking u dude

Mathematics

1 answer:

murzikaleks [220]3 years ago

8 0

Answer:

.............................

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A. Name the smallest angle and the largest angle of the following triangles: F 6 S 5 1. A J 5.03 7 5 G B H 5.1 2. D 4 3 E

I am Lyosha [343]

Answer:

(smallest, largest) = (B, C), (D, E), (G, J)

Step-by-step explanation:

Side lengths of a triangle are proportional to the sine of the opposite angle. In a triangle, a larger angle will always have a larger sine. That means the smallest angle is opposite the shortest side, and the largest angle is opposite the longest side.

The shortest side is 5 units, opposite angle B. The longest side is 7 units, opposite angle C.

smallest angle: B
largest angle: C

The shortest side is 3 units, opposite angle D. The longest side is 5 units, opposite angle E.

smallest angle: D
largest angle: E

The shortest side is 5 units, opposite angle G. The longest side is 5.1 units, opposite angle J.

smallest angle: G
largest angle: J

_____

Additional comment

Angles in a triangle may exceed 90°. The sine function actually is decreasing for angles above 90°. However, since the total of angles of a triangle must be exactly 180°, there cannot be two angles in a triangle such that the smaller angle has the larger sine.

6 0

2 years ago

How do I find a common denominator for a complex faction when one denominator is 2m-1 and the other is 2m+5

ratelena [41]

The best option is to multiply them together, there would be no value that is common between the two, (2m - 1)(2m + 5), and leave it in this form.

However, there is a second method.
Notice how between the two sits a difference of 6.
You can manipulate one of the fractions to render the same denominator and work from there.

3 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

Please help me!!! 10 points and I will count u as brainliest

Lena [83]

Answer:

F or H

Step-by-step explanation:

not J because that is negative, not G because that means none, so either H or F. I'm not going to give you the whole answer but another hint is look at the circles are they circled or non circlircled. I eliminated two answers for you.

remember what your teachers saids: what does it mean for the circlies to be colored or blank.

good luck!

5 0

3 years ago

Which one is the word form of 3 x 5/6

Maslowich

Three times five sixths

3 0

3 years ago

Read 2 more answers