3 years ago

I need help with radical expression and need to show my work

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

6 0

See the attached picture for the answer and steps:

You might be interested in

Which of the following statements about JK and JL is true?

bulgar [2K]

Answer:

Step-by-step explanation:

5 0

3 years ago

Câu 99: Cho hàm số f x  xác định trên ℝ , bảng biến thiên của hàm số f x   như sau.

OLga [1]

Answer:

i doesn,t understand this language

4 0

3 years ago

Find each quotient using long division. Just type the number answer in the blank.<br> Pleaseeee help

sergey [27]

Answer: 321 for #1 and 52 for #2

Step-by-step explanation: I divided and got 321 for #1 and 52 for #2.

8 0

3 years ago

The segments shown below could form a triangle,

Brrunno [24]

Answer:

False

Step-by-step explanation:

The lengths do not adhere to the triangle inequality theorem. Which states that the sum of the side lengths of any 2 sides of a triangle must exceed the length of the third side.

4 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