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2 years ago

What are the values of x for which the denominator is equal to zero for y=x^2+3/x^2+2

Mathematics

1 answer:

Contact [7]2 years ago

7 0

Answer:

when x = ± $\sqrt{2}$

Step-by-step explanation:

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Select all expressions that can be subtracted from 9x to result in the expression 3x+5.

Arte-miy333 [17]

<h2>Answer:</h2><h2>The expressions that can be subtracted from 9x to result in the expression 3x+5 is 6x - 5</h2>

Step-by-step explanation:

To find the expression that can be subtracted from 9x to result in the expression 3x+5.

Let us consider the expression, a - b = c

Here a = 9x

the result, c = 3x + 5

b = ?

Substitute the values in the above equation, we get

9x - b = 3x + 5

9x - (3x + 5) = b

b = 6x - 5

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3 years ago

Can anyone help me on solving equations with rational numbers?

dangina [55]

I'm learning the exact same thing so give me some question and I can see what I can do!

8 0

3 years ago

I need help with this ^

Nady [450]

<h3>Answer is 20 </h3>

<h2>Please mark me as Brainliest ......</h2>

5 0

3 years ago

(b) dy/dx = (x - y+ 1)^2

Elanso [62]

Substitute $v(x)=x-y(x)+1$ , so that

$\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}$

Then the resulting ODE in $v(x)$ is separable, with

$1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx$

On the left, we can split into partial fractions:

$\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx$

Integrating both sides gives

$\dfrac{\ln|1-v|+\ln|1+v|}2=x+C$

$\dfrac12\ln|1-v^2|=x+C$

$1-v^2=e^{2x+C}$

$v=\pm\sqrt{1-Ce^{2x}}$

Now solve for $y(x)$ :

$x-y+1=\pm\sqrt{1-Ce^{2x}}$

$\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}$

3 0

3 years ago

I am not sure if i did this right she wants us to distribute and get only one x on one side and find the answer.

umka21 [38]

I got x=18

3(x-4)-2(3x+4)=4(3-x)+5x+4

3x-12-6x-8=12-4x+5x+4

3x-20=16+x

2x=36x

x=18

5 0

2 years ago