3 years ago

What is the distance between (-5, -6) and (-3,-8)?

Mathematics

1 answer:

Mashcka [7]3 years ago

3 0

Answer:-2 and 2

Step-by-step explanation:

You might be interested in

Write the number 33×10^-3 in scientific notation

cluponka [151]

Answer:

Step-by-step explanation:

33×10^-3 is equivalent to 33/1000, or 0.033.

This, in turn, is equivalent to 3.3×10^(-2) (which is in scientific notation)

4 0

3 years ago

Charlie is flying a kite one afternoon and steps on the end of the string to have his

Butoxors [25]

Answer:

roubd it to 20 and you will get 679.90

7 0

3 years ago

Find each sum or difference. Write in simplest fraction form. Pamela ran 7/10 mile on Tuesday and 3/8 mile on Thursday. How much

Jlenok [28]

She ran 13/40 of a mile more on Tuesday

8 0

3 years ago

Plsssssss help i need it.

riadik2000 [5.3K]

I would say decreasing!

7 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