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3 years ago

EASY MATH PROBLEM!!!!

Mathematics

1 answer:

Mkey [24]3 years ago

4 0

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With a detailed explanation from the internet 1/(x-5)+3/(x+2)=4

irga5000 [103]

Answer:

$\boxed{ \sf x=\dfrac{4\pm\sqrt{43}}{2}}$

Explanation:

$\rightarrow \sf \dfrac{1}{x-5}+\dfrac{3}{x+2}=4$

make the denominators same

$\rightarrow \sf \dfrac{1(x+2)}{(x-5)(x+2)}+\dfrac{3(x-5)}{(x+2)(x-5)}=4$

join the fractions together

$\rightarrow \sf \dfrac{x+2+3x-15}{(x-5)(x+2)}=4$

cross multiply

$\rightarrow \sf x+2+3x-15=4(x-5)(x+2)$

simplify

$\rightarrow \sf 4x-13=4x^2-12x-40$

group the variables

$\rightarrow \sf 4x^2-16x-27 = 0$

use quadratic formula

$\rightarrow \sf x = \dfrac{-\left(-16\right)\pm \sqrt{\left(-16\right)^2-4\cdot \:4\left(-27\right)}}{2\cdot \:4}$

simplify the following

$\rightarrow \sf x=\dfrac{4\pm\sqrt{43}}{2}$

7 0

2 years ago

Read 2 more answers

33=-3 (3-2) solve the equation please I need help

hammer [34]

Where is the variable?

6 0

3 years ago

An arithmetic series consists of consecutive integers that are multiples are of 4. What is the sum of the first nine terms of th

Mrac [35]

Answer: The answer is 144.

Step-by-step explanation: Let 'a' be the first term and 'd' be the common difference of the given arithmetic progression (A.P.).

According to the given information, A.P. is {0, 4, 8, 12, 16, . . .}, i.e., a=0 and d=4.

The sum of first n terms is given by

$S_n=\dfrac{n}{2}\{2a+(n-1)d\}.$

So, the sum of first 9 terms is

$S_9=\dfrac{9}{2}\{2\times 0+(9-1)\times4\}=\dfrac{9}{2}(8\times 4)=144.$

Thus, the required sum is 144.

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%7B%5Chuge%7B%5Cunderline%7B%5Csmall%7B%5Cmathbb%7B%5Cpink%7BREFER%20%5C%20TO%20%5C%20THE%20%5

marusya05 [52]

Answer:

See below

(B) and (C) are correct.

Step-by-step explanation:

We have the following limit

$\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} }, \forall x>0$

I am not sure about methods concerning the quotient, but in this type of question I would try to convert this limit into integration.

Considering the numerator, we have

$(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right) = \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)$

- I didn't forget about $n^n$

Considering the denominator, we have

$(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right) = \prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)$

- I didn't forget about $n!$

Therefore,

$\left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} } = \left(\dfrac{n^2 \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)}{ n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)} \right)$

$= \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

Now we have

$\lim \limits_{n\rightarrow \infty} \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}, \forall x>0$

This is just the notation change so far.

What I want to do here is apply definite integrals using Riemann Integrals (We will write the limit as an definite integral). A nice way to do it is using logarithms. Therefore, we can apply the natural logarithm in both sides.

Now, recall two properties of logarithms:

$\boxed{\log_a mn = \log_a m + \log_a n}$

$\boxed{\log_a m^p = p\log_a m}$

$\boxed{\log_a \left(\dfrac{m}{n} \right) = \log_a m- \log_a n}$

Thus,

$\ln f(x) = \lim \limits_{n\rightarrow \infty} \ln \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n}\ln\left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln \left(n^n \prod_{k=1}^n \left(x+\dfrac{n}{k} \right) \right)-\ln \left( n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right) \right) \right]$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln n^n + \prod_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)-\ln n! -\prod_{k=1}^n \ln\left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Considering

$\lim \limits_{n\rightarrow \infty} \frac{x}{n} (\ln n^n - \ln n! ) = \lim \limits_{n\rightarrow \infty} \frac{x}{n} (n\ln n - \ln n! )= \lim \limits_{n\rightarrow \infty} \frac{x \cdot\ln\frac{n^n}{n!} }{n}$

Using Stirling's formula

$\dfrac{n^n}{n!}\underset{\infty}{\sim} \dfrac{n^n}{\sqrt{2n \pi}\left(\dfrac{n}{e}\right)^n}=\dfrac{e^n}{\sqrt{2n \pi}}$

then

$\ln\left(\frac{n^n}{n!}\right)\underset{\infty}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)$

$\implies \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$

This shows our limit equals 1 as $\frac{\log(2\pi n)}{2n} \rightarrow 0$ and $\ln(e)=1$

Employing a Riemann sum in the main limit, we have

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[ \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) - \sum_{k=1}^n\ln \left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Now dividing the terms inside the parenthesis by $\dfrac{n}{k}$ in $\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)$

we have

$\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) = \sum_{k=1}^n \ln \left(\frac{kx}{n} +1\right)$

Now dividing the terms inside the parenthesis by $\dfrac{n^2}{k^2}$ in $\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)$

we have

$\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right) = \sum_{k=1}^n \ln \left(\frac{(kx)^2}{n^2} +1\right)$

Therefore

$= \frac xn\sum_{k=1}^n\ln\dfrac{z+1}{z^2+1}$

for $\dfrac{kx}{n} = z$

Using Riemann Integral,

$\lim \limits_{n\rightarrow \infty} \int_0^1\ln\frac{z+1}{z^2+1}dz$

From

$\frac{f'(x)}{f(x)}=\ln\frac{z+1}{z^2+1}$

We can see that the function is increasing for , but because of the denominator, it is negative for .

Therefore,

(A) is false because $\dfrac{1}{2} < 1$

(B) is true because

(D) is false, just consider $\ln\frac{z+1}{z^2+1}$ for $z=1$ and $z=2$

7 0

3 years ago

Find x of this triangle.

julsineya [31]

Answer:

x=18

Step-by-step explanation:

First, using the Pythagorean theorem, you can make the equation 10^2+y^2=26^2. You can solve this to get y=24. Another way to solve for the value of y is to use your Pythagorean triples! (in this case, all the values are double the values of the Pythagorean triple 5, 12, 13).

Next, you can use the Pythagorean theorem again to get the equation 24^2+x^2=30^2. Solving it, you would get x=18.

Hope this helps!

3 0

3 years ago