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Troyanec [42]
3 years ago
15

Qtr is a right triangle TQR is a right triangle QS is an altitude which statements are true

Mathematics
1 answer:
Artemon [7]3 years ago
4 0

Answer:

the answer is 1and 3 that's right

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PLEASE HELPP!! WILL GIVE BRAINLIEST + 25 POINTS!!
enyata [817]

Answer:

x = -9/4

Step-by-step explanation:

−2(x + 1/4) + 1 = 5 ​

Subtract 2 from both sides.

−2(x + 1/4) = 4 ​

Divide both sides by -2.

x + 1/4 = -2

Subtract 1/4 from both sides.

x = -2 1/4

x = -9/4

4 0
3 years ago
What is 3,989.23655 rounded to the nearest one thousandth?
Xelga [282]
The 6 is in the thousandth place so you round with that being the last number
since we have a 55 after the 6, we round up
so we get
3,989.237
8 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
In a class of students, the following data table summarizes
Ksenya-84 [330]

Answer:

Step-by-step explanation:

you have t as a total amount of the whole class.

f= did not pass and you place the total number next to it like this;  f/t represents that group. You continue this in your workings.

f/t = probability failed.

p/t = probability passed

c/t = probability completed

dc/t = probability did not complete

and draw a tree.

The random would be either of these, we know there s a possibility of 1/4 but the odds stay central. We look for a new way after this data is converted as a  fraction and show these fractions as your answer.

4 0
3 years ago
6 - 2x = 6x - 10x + 4
yarga [219]

6 - 2x = 6x - 10x + 4

6 - 2x =  - 4x + 4

6 =  - 4x + 4 + 2x

6 =  - 2x + 4

- 2x + 4 = 6

- 2x = 6 - 4

- 2x = 2

x =  \frac{ \:  \:  \: 2}{ - 2}

x =  - 1

Therefore , the value of x = -1

6 0
3 years ago
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