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4 years ago

Is 0.3 equal to 1/3, or 3/10?

Mathematics

2 answers:

Stolb23 [73]4 years ago

8 0

Answer: 3/10

3/10 = 0.3

xenn [34]4 years ago

4 0

Answer:

It is equal to 3/10.

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What is 11.3 times 4.3

s2008m [1.1K]

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48.59

Step-by-step explanation:

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4 years ago

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Examine the following graph of y=1/4x+2.

choli [55]

Answer: 9/4

Step-by-step explanation:

The slope-intercept form is y = m x + b , where m is the slope and b is the y-intercept. y = m x + b Find the values of m and b using the form y = m x + b . m = 1/4 b = 2 The slope of the line is the value of m , and the y-intercept is the value of b . Slope: 1/4 y-intercept: 2 Any line can be graphed using two points. Select two x values, and plug them into the equation to find the corresponding y values. Choose 0 to substitute in for x to find the ordered pair. y = 2 Choose 1 to substitute in for x to find the ordered pair. Tap for fewer steps... Replace the variable x with 1 in the expression. f ( 1 ) = 1 4 + 2 Simplify the result. The y value at x = 1 is 9/4 . y = 9/4

3 0

4 years ago

<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:

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