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kenny6666 [7]
3 years ago
8

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:

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

You might be interested in
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.

__

<h3>ΔABC</h3>

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

  • smallest angle: B
  • largest angle: C
<h3>ΔDEF</h3>

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

  • smallest angle: D
  • largest angle: E
<h3>ΔGHJ</h3>

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

_____

<em>Additional comment</em>

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 <em>cannot be two angles in a triangle such that the smaller angle has the larger sine</em>.

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 <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
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
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