Ask question

Ask question

All categories

Aleonysh [2.5K]

3 years ago

5

Solve the equation 5/x+3 + 4/x+2 = 2

1 answer:

Bess [88]3 years ago

6 0

Answer:

5 + 4 = 9

x × x = X2

3 + 2 = 5

9/x + 5 = 2

You might be interested in

In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is

zheka24 [161]

Answer:

$f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

Step-by-step explanation:

Let:

$M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y$

This is and exact equation, because:

$\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}$

So, define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)$

The solution will be given by:

$f(x,y)=C_1$

Where C1 is an arbitrary constant

Integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y):

$f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)$

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}$

Substitute into $\frac{\partial f(x,y)}{\partial y} =N(x,y)$

$2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y$

Integrate $\frac{dg(y)}{dy}$ with respect to y:

$g(y)=\int\ {8y} \, dy =4y^2$

Substitute g(y) into f(x,y):

$f(x,y)=2x^2+4y^2+2xy$

The solution is f(x,y)=C1

$f(x,y)=2x^2+4y^2+2xy=C_1$

Solving y using quadratic formula:

$y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )$

4 0

4 years ago

Which of the following pairs of numbers has at least common multiple of 36?

Aleks [24]

0,3,6,9,12,15,18,21,24,27,30,33,36
0,12,24,36

3 0

3 years ago

Applying properties of Exponents in exercise,use the properties of exponents to simplify the expression.See example 1.

evablogger [386]

Answer: a) 15625, b) 0.2, c) 625, d) 0.008.

Step-by-step explanation:

Since we have given that

a) $(5^2)(5^3)$

As we know that

$a^m\times a^n=a^{m+n}$

So, it becomes,

$5^2\times 5^3=5^{2+3}=5^6=15625$

(b) $(5^2)(5^{-3})$

So, it becomes,

$5^2\times 5^{-3}=5^{2-3}=5^{-1}=\dfrac{1}{5}=0.2$

(c) $(5^2)^2$

As we know that

$(a^m)^n=a^{mn}$

So, it becomes,

$5^{2\times 2}=5^4=625$

(d) $5^{-3}$

$a^{-m}=\dfrac{1}{a^m}\\\\5^{-3}=\dfrac{1}{5^3}=\dfrac{1}{125}=0.008$

Hence, a) 15625, b) 0.2, c) 625, d) 0.008.

4 0

3 years ago

What is the approximate value of <br> f(x)= 74/1+2e^-1.5x

Anton [14]

Answer: Oblique Asymptotes: y = 2 x/ e^1.5

Step-by-step explanation: Find the asymptotes

brainliest or a thank you pls <33 :)

4 0

3 years ago

When the outliers are removed how does the mean change? Dot plot with one on 18, 2 on 38, one on 40, 3 on 44, one on 46, one on

Vitek1552 [10]

18,38,38,40,44,44,44,46,48....
the mean with the outlier (18) included is : 360/9 = 40

38,38,40,44,44,44,46,48
the mean with the outlier (18) excluded is : (360 - 18) / 9 = 342/9 = 38

so when the outlier is removed, the mean is 2 points less

4 0

3 years ago

Read 2 more answers