All categories

3 years ago

Witch is grater 3/4 or 11/16 ??

1 answer:

5 0

<h3>Hey there! </h3><h3> $\frac{3}{4}$ → $0.75$ → $0.75$ %</h3><h3> $\frac{11}{16}$ → $0.6875$ → $0.68$ % </h3><h3>So, the one that is $GREATER$ is: $\frac{3}{4}$ because $75$ % is bigger than</h3><h3> $68$ %</h3><h3>Good luck on your assignment and enjoy your day! </h3><h3>~ $LoveYourselfFirst:)$ </h3>

You might be interested in

<img src="https://tex.z-dn.net/?f=%20%20%5Cdisplaystyle%20%5Cint%20%5Climits_%7B0%7D%5E%7B%20%5Cfrac%7B%20%5Cpi%7D%7B2%7D%20%7D%

murzikaleks [220]

Let $x = \arcsin(y)$ , so that

$\sin(x) = y$

$\tan(x)=\dfrac y{\sqrt{1-y^2}}$

$dx = \dfrac{dy}{\sqrt{1-y^2}}$

Then the integral transforms to

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_{y=\sin(0)}^{y=\sin\left(\frac\pi2\right)} \frac{y}{\sqrt{1-y^2}} \ln(y) \frac{dy}{\sqrt{1-y^2}}$

$\displaystyle \int_{x=0}^{x=\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy$

Integrate by parts, taking

$u = \ln(y) \implies du = \dfrac{dy}y$

$dv = \dfrac{y}{1-y^2} \, dy \implies v = -\dfrac12 \ln|1-y^2|$

For 0 < y < 1, we have |1 - y²| = 1 - y², so

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = uv \bigg|_{y\to0^+}^{y\to1^-} + \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

It's easy to show that uv approaches 0 as y approaches either 0 or 1, so we just have

$\displaystyle \int_0^1 \frac{y}{1-y^2} \ln(y) \, dy = \frac12 \int_0^1 \frac{\ln(1-y^2)}{y} \, dy$

Recall the Taylor series for ln(1 + y),

$\displaystyle \ln(1+y) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n y^n$

Replacing y with -y² gives the Taylor series

$\displaystyle \ln(1-y^2) = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n (-y^2)^n = - \sum_{n=1}^\infty \frac1n y^{2n}$

and replacing ln(1 - y²) in the integral with its series representation gives

$\displaystyle -\frac12 \int_0^1 \frac1y \sum_{n=1}^\infty \frac{y^{2n}}n \, dy = -\frac12 \int_0^1 \sum_{n=1}^\infty \frac{y^{2n-1}}n \, dy$

Interchanging the integral and sum (see Fubini's theorem) gives

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy$

Compute the integral:

$\displaystyle -\frac12 \sum_{n=1}^\infty \frac1n \int_0^1 y^{2n-1} \, dy = -\frac12 \sum_{n=1}^\infty \frac{y^{2n}}{2n^2} \bigg|_0^1 = -\frac14 \sum_{n=1}^\infty \frac1{n^2}$

and we recognize the famous sum (see Basel's problem),

$\displaystyle \sum_{n=1}^\infty \frac1{n^2} = \frac{\pi^2}6$

So, the value of our integral is

$\displaystyle \int_0^{\frac\pi2} \tan(x) \ln(\sin(x)) \, dx = \boxed{-\frac{\pi^2}{24}}$

6 0

3 years ago

What number is 35% of 22?

ankoles [38]

7.7 hope this helped

3 0

4 years ago

Read 2 more answers

Suppose you want to have $400,000 for retirement in 35years. Your account earns 4% interest.

Sedaia [141]

$\begin{equation*} \$437.76 \end{equation*}$

1) This is a question for Annuity usage. Note that we'll need the following formula below:

$\begin{gathered} P=\frac{A(\frac{r}{n})}{\lbrack(1+\frac{r}{n})^{nt}-1\rbrack} \\ P=\frac{400000(\frac{0.04}{12})}{\lbrack(1+\frac{0.04}{12})^{12\cdot35}-1\rbrack} \\ P=\frac{400000\cdot\frac{0.04}{12}}{\left(1+\frac{0.04}{12}\right)^{35\cdot\:12}-1} \\ P=\frac{400000\cdot\frac{0.04}{12}}{1.0033^{420}-1} \\ P=437.76563...\approx\$437.76 \end{gathered}$

Note that P is the value you'll need to deposit each month so that in 35 years you'll have $400,000 for your retirement.

6 0

1 year ago

On Monday, the students in Mr. Carson’s class worked in pairs. On Tuesday, the students worked in groups of 3. On Wednesday, the

qwelly [4]

The smallest number of students in the class is the smallest number that is divisible by 2, 3, and 4. You need the least common multiple of 2, 3, and 4.
This is like finding the least common denominator.

Find the prime factors of the three numbers:
2 = 2
3 = 3
4 = 2^2

To find the LCM you need common factors, and not common factors with the larger exponent.

Now single factor is common to all numbers.
There is 2 and 2^2, so use 2^2 since it has the larger exponent.
There is also a 3, so use the 3.

LCM = 2^2 * 3 = 4 * 3 = 12

Mr. Carson's class may have 12 students. 12 is the smallest number of students it can have.

3 0

4 years ago

I NEED HELP PLEASE. THANK YOU

vova2212 [387]

Answer:

Given

Distributive property

Subtraction property of equality

Simplify

Subtraction property of equality

Simplify

Division property of equality

Simplify

Step-by-step explanation:

If you want an explanation I would be happy to provide you with one, but I do not want to have to explain the entire thing right here because it would be really long.

7 0

3 years ago