All categories

3 years ago

3 + 1/3 equals b - 23 State the problem

Mathematics

1 answer:

Viefleur [7K]3 years ago

5 0

Answer:

$b=\frac{79}{3}$ or $b=26.33$

Step-by-step explanation:

Assuming the question to be finding the value of b.

$3+\frac{1}{3}=b-23$

Solution:

Rewrite the equation as.

$b-23=3+\frac{1}{3}$

first we solve right side of the equation.

$b-23=\frac{3\times 3+ 1}{3}$

$b-23=\frac{9+ 1}{3}$

$b-23=\frac{10}{3}$

Add 23 both side of the equation.

$b-23+23=\frac{10}{3}+23$

$b=\frac{10+23\times 3}{3}$

$b=\frac{10+69}{3}$

$b=\frac{79}{3}$

$b=26.33$

Therefore, the value of $b=\frac{79}{3}$ or $b=26.33$

You might be interested in

Julie keeps rabbits and hamsters as pets she has four rabbits and an odd number of hamsters she has more hamsters than rabbits t

son4ous [18]

Answer:

H is greater than 4, less than 6, and is odd. H, therefore, must be 5.

Step-by-step explanation:

Let's write down what we know:

Julie keeps Rabbits ("R") and Hamsters ("H") as pets.

R=4

She has more H than R, so:

H>4

and H is odd.

The total of R and H is less than 10, so

R+H<10

we can substitute in for R:

4+H<10

and now we can solve for H:

H<10−4=6

So we know that H is greater than 4, less than 6, and is odd. H, therefore, must be 5.

3 0

3 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:

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