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3 years ago

A partial cylinder lies on its side. The bases are a 90° sector of a circle. What is the exact volume of the partial cylinder?

Mathematics

2 answers:

LenKa [72]3 years ago

8 0

Answer:

180%

Step-by-step explanation:

777dan777 [17]3 years ago

8 0

Answer:

160piein^3

Step-by-step explanation:

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Graph the image of the figure after a dilation with a scale factor of 2 centered at (-7, - 2).

tigry1 [53]

Answer:

Sorry this is late!

Step-by-step explanation:

8 0

3 years ago

When rabbits were introduced to the continent of Australia they quickly multiplied and spread across the continent since there w

Lady bird [3.3K]

Answer:

a. $y=6(1.7472)^x$

b. $y=6e^{0.558t}$

c.13.3 months

Step-by-step explanation:

a.-Given the first term at $t_0$ is 6 and the second term at $t_3$ is 32.

-Let's take rabbit population as a function of time to be

$y=ab^x$

where y is the population at time x and a the initial population at $t_0\\$

#We substitute our values to calculate the value of the constant b:

$y_x=ab^x\\\\y_3=ab^3\\\\32=6b^3\\\\b=1.472$

#Replace b in the population function:

$y=ab^x, b=1.7472,a=6\\\\\therefore y=6(1.7472)^x$

Hence, the regression for the rabbit population as a function of time x is $y=6(1.7472)^x$

b. The exponential function in terms of base $e$ is usually expressed as:

$A=A_0e^{kt}$

Where:

$A_0$ -is the initial population at $t_o$

$A$ -is the population at time t.

$k-$ is the exponential growth constant.

$e-$ the exponent

Our function in terms of base exponent is rewritten as:

$y=A_0e^{kt}$

#Substitute with actual figures to solve for t:

$y=A_0e^{kt}, y=32, xt=3, A_0=6\\\\32=6e^{3k}\\\\3k=In (32/6)\\\\k=0.5580$

Hence, the regression equation in terms of base e is $y=6e^{0.558t}$

c. We substitute y with any number higher than 10,000 to estimate the time for the rabbits to exceed 10,000.

-We know that $y=6e^{0.558t}.$

Therefore we calculate t as(take y=10001):

$y=6e^{0.558t}, y=10001\\\\10001=6e^{0.558t}\\\\1666.8333=e^{0.558t}\\\\0.558t= In 1666.8333\\\\t=13.2951$

Hence, it takes approximately 13.3 months for the population to exceed 10000

3 0

4 years ago

A man traveled a certain number of miles by automobile and then nine times as far by airplane. His total trip was 600 miles in l

Alex17521 [72]

The miles he traveled by plane is nine times greater than the miles traveled by automobile. If the total is 600 miles, then the total distance traveled by automobile would be 60 miles, making the total traveled by airplane 540 miles.

4 0

3 years ago

Sherry claims that the expression 1x will always be equivalent to a repeating decimal whenever x is an odd number greater than 1

Zinaida [17]

This question is not complete

Complete Question

Sherry claims that the expression 1/x will always be equivalent to a repeating decimal whenever x is an odd number greater than 1.

Which of these values of x will prove Sherry's claim is false?

Answer:

When x = 5

Step-by-step explanation:

Sherry claims that the expression 1/x will always be equivalent to a repeating decimal whenever x is an odd number greater than 1.

Examples of odd numbers greater than 1 : 3, 5, 7, 9, 11 ....

We would put these odd numbers to test

a) When x = 3

= 1/3 = 0.3333333333

b) When x = 5

= 1/5 = 0.2

c) When x = 7

= 1/7 = 0.142857142

d) When x = 9

= 1/9 = 0.1111111111

e) When x = 11

= 1/11 = 0.0909090909

From the above calculation, we can see that the only odd number greater than 1 that will prove Sherry's theory wrong is when x = 5

Therefore, the value of x that will prove Sherry's claim is false is when x = 5

3 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.