Ask question

Ask question

All categories

3 years ago

10

A rectangular pool is 6 meters wide and 8 meters long. If you swim diagonally across the pool, how many meters would you be swim

ming?

1 answer:

Vitek1552 [10]3 years ago

4 0

If you swim diagonally across the rectangular pool, the distance you swim is 10 meters.

<u>Given the following data:</u>

Width of rectangle = 6 meters

Length of rectangle = 8 meters

To determine the distance you swim in meters, we would apply Pythagorean's theorem since the width is along the x-axis while the length is along the y-axis.

Note: The diagonal side of the rectangular pool represents the hypotenuse.

Mathematically, Pythagorean's theorem is given by the formula:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

Substituting the given parameters into the formula, we have;

$Hypotenuse^2 = 6^2 + 8^2\\\\Hypotenuse^2 = 36+64\\\\Hypotenuse^2 = 100\\\\Hypotenuse = \sqrt{100}$

Hypotenuse = 10 meters.

Read more here: brainly.com/question/18890335

You might be interested in

1. Find the equation of a line passing

lina2011 [118]

1. m= 1/2
2. x=-2

hope this helps!

8 0

3 years ago

PLEASE HELP<br> Complete the table by computing the amount in the account at the end of each period.

skad [1K]

Answer:

See below

Step-by-step explanation:

Initial Principal P₀ = 1000
End of first period = 1000 + 1000*0.02/4 = 1005
End of second period = 1005 + 1000*0.02/4 = 1010
End of third period = 1010 + 1000*0.02/4 = 1015
End of fourth period = 1015 + 1000*0.02/4 = 1020

8 0

3 years ago

I have calculus problems that I need help with.

aleksklad [387]

a. Note that $f(x)=x^ne^{-2x}$ is continuous for all $x$ . If $f(x)$ attains a maximum at $x=3$ , then $f'(3) = 0$ . Compute the derivative of $f$ .

$f'(x) = nx^{n-1} e^{-2x} - 2x^n e^{-2x}$

Evaluate this at $x=3$ and solve for $n$ .

$n\cdot3^{n-1} e^{-6} - 2\cdot3^n e^{-6} = 0$

$n\cdot3^{n-1} = 2\cdot3^n$

$\dfrac n2 = \dfrac{3^n}{3^{n-1}}$

$\dfrac n2 = 3 \implies \boxed{n=6}$

To ensure that a maximum is reached for this value of $n$ , we need to check the sign of the second derivative at this critical point.

$f(x) = x^6 e^{-2x} \\\\ \implies f'(x) = 6x^5 e^{-2x} - 2x^6 e^{-2x} \\\\ \implies f''(x) = 30x^4 e^{-2x} - 24x^5 e^{-2x} + 4x^6 e^{-2x} \\\\ \implies f''(3) = -\dfrac{486}{e^6} < 0$

The second derivative at $x=3$ is negative, which indicate the function is concave downward, which in turn means that $f(3)$ is indeed a (local) maximum.

b. When $n=4$ , we have derivatives

$f(x) = x^4 e^{-2x} \\\\ \implies f'(x) = 4x^3 e^{-2x} - 2x^4 e^{-2x} \\\\ \implies f''(x) = 12x^2 e^{-2x} - 16x^3e^{-2x} + 4x^4e^{-2x}$

Inflection points can occur where the second derivative vanishes.

$12x^2 e^{-2x} - 16x^3 e^{-2x} + 4x^4 e^{-2x} = 0$

$12x^2 - 16x^3 + 4x^4 = 0$

$4x^2 (3 - 4x + x^2) = 0$

$4x^2 (x - 3) (x - 1) = 0$

Then we have three possible inflection points when $x=0$ , $x=1$ , or $x=3$ .

To decide which are actually inflection points, check the sign of $f''$ in each of the intervals $(-\infty,0)$ , $(0, 1)$ , $(1, 3)$ , and $(3,\infty)$ . It's enough to check the sign of any test value of $x$ from each interval.

$x\in(-\infty,0) \implies x = -1 \implies f''(-1) = 32e^2 > 0$

$x\in(0,1) \implies x = \dfrac12 \implies f''\left(\dfrac12\right) = \dfrac5{43} > 0$

$x\in(1,3) \implies x = 2 \implies f''(2) = -\dfrac{16}{e^4} < 0$

$x\in(3,\infty) \implies x = 4 \implies f''(4) = \dfrac{192}{e^8} > 0$

The sign of $f''$ changes to either side of $x=1$ and $x=3$ , but not $x=0$ . This means only $\boxed{x=1}$ and $\boxed{x=3}$ are inflection points.

4 0

1 year ago

Read 2 more answers

85 quarts equals how many pints?

Cloud [144]

85 Quarts equals 170 Pints Because:
1 Quart = 2 Pints
85*2=170 Pints

8 0

4 years ago

Read 2 more answers

Yall i aint tryna make fun but one time i was in the school hallway and this slow kid was walking like a zombie and my best frie

Karolina [17]

Answer:

PLS- not a zombie apocolypse

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers