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Sophie [7]
3 years ago
5

Before sunrise, the temperature was 22.4°F below zero. By noon, it had risen 12.8°F. What expression can be used to find the

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
8 0

Answer:

-22.4+12.8

Step-by-step explanation:

If the temperature started at 22.4 below zero you would write that as -22.4 because it had risen 12.8 degrees you would add those 12.8 degrees to -22.4

MrMuchimi3 years ago
6 0

Answer:-22.4+12.8

Step-by-step explanation:

You might be interested in
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
A segment can have more than one bisector.<br> True<br> False
insens350 [35]
<span>Can a Segment have more than one bisector. Yes A segment can have more than one bisector. For every line segment, there is one perpendicular bisector that passes through the midpoint. There are infinitely many bisectors, but only one perpendicular bisector for any segment.</span>
4 0
3 years ago
A. Write the equation ax + b = c in terms of x.
Oksanka [162]

Answer:

x=c/a-b/a

Step-by-step explanation:

ax+b=c

1) Subtract b from both sides:

ax=c-b

2) Divide both sides by a:

x=c/a-b/a

5 0
3 years ago
Read 2 more answers
In ΔCDE, the measure of ∠E=90°, the measure of ∠D=22°, and CD = 80 feet. Find the length of EC to the nearest tenth of a foot.
chubhunter [2.5K]

Answer:

30

Step-by-step explanation:

5 0
3 years ago
When the full season tickets first went on sale,2,000 Full season tickets sold for Section N. Two weeks after the tickets first
slega [8]

Image showing the arena ticket prices is missing, so i have attached it.

Answer:

$10000 more money was spent when the tickets first went on sale than after the first 2 weeks

Step-by-step explanation:

We are told that when the full season tickets first went on sale, that 2000 full season tickets were sold for section N.

Now, from the area ticket prices table attached, we can see that full season tickets for Section N costs $20

Thus, amount of money spent when the tickets first went on sale = 2000 × 20 = $40000

We are told that 2 weeks after the tickets first went on sale, they sold 1500 tickets. Thus, amount spent after 2 weeks release = 1500 × 20 = $30000

Difference in amount spent at the beginning and after 2 weeks = $40000 - $30000 = $10000

Thus, $10000 more money was spent when the tickets first went on sale than after the first 2 weeks

6 0
3 years ago
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