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3 years ago

Select all the sets of numbers that are possible values for x in the inequality, x<-3.

Mathematics

1 answer:

Solnce55 [7]3 years ago

6 0

Answer: 2 and 3

Step-by-step explanation:

all (-9,-7,-5) and (-33,-22,-11) are negative numbers less than -3. So those answers are possible answers. ps i had this question on my test too.

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Susan tips at the rate of $75 to 5 waiters. How many waiters will she be tipping,if she can afford $90 tip?

mash [69]

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3 years ago

3packs of soda is 10 dollars less than 5 packs of soda equation

RideAnS [48]

Ok. so lets call the cost of each soda x. now, 3 packs of soda is 10 less than five packs of soda. so the price of 3 packs of soda is 3x, the price of five packs is 5x. now, it says that 3 packs is 10 less than five packs. that means that 3x=5x-10.
now, to find the price of a pack of soda, we have to isolate x.
add 10 to both sides to get 3x+10=5x. now, we subtact 3x on both sides to get 10=2x. divide by 2 on both sides, x=5. a pack of soda is 5 dollars

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3 years ago

Which statements describe a parallelogram that must be a square? Select each correct answer.

Oxana [17]

That would be B, C and E.

A and D could be a rhombus.

8 0

3 years ago

Read 2 more answers

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

Solve for x. Round to the nearest tenth if necessary.

timofeeve [1]

Answer: 41.1

Step-by-step explanation:

To find the value of x, we need to use SOH-CAH-TOA. SOH-CAH-TOA is since, cosine, and tangent. The O, A, H stands for opposite, adjacent, and hypotenuse respectively.

Looking at the figure, we see 47°. The labelled angles are adjacent and the hypotenuse. Therefore, we use CAH or cosine.

$cos(47)=\frac{28}{x}$ [multiply both sides by x]

$xcos(47)=28$ [divide both sides by cos(47)]

$x=\frac{28}{cos(47)}$

When we plug that into a calculator, we get x=41.1.

5 0

3 years ago

Read 2 more answers