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3 years ago

What is 250 as a power of 4

Mathematics

1 answer:

belka [17]3 years ago

8 0

Answer:

3,906,250,000

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List all the points that are on the same plane as point V

topjm [15]

I believe the points should be as follows : X, S, Y, and U.

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3 years ago

Which function has the same y-intercept as the function y equals 2/3 x - 3

aleksley [76]

We have the function $y = \frac{2}{3} x -3$ and we want to find a function that has the same y-intercept than the previous function.

First, let's find the y-intercept by subtituting 0 for 'x'.

$y = \frac{2}{3} (0) -3 = -3$

Now that we found that y-intercept =-3, any lineal function of the type: $y = ax - 3$ will have the same y-intercept. Where 'a' can take all the real values.

Also, any quadratic function of the type: $y=ax^{2} + bx - 3$ will have the same y-intercept. Where 'a' and 'b' can take all the real values.

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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

Simplify to a single power of 4:<br> 4^7/4^6

Temka [501]

Step-by-step explanation:

$\frac{ {4}^{7} }{ {4}^{6} }$

= 4

.......

4 0

3 years ago

Read 2 more answers

Can i get help please i don't know this

scoundrel [369]

I think it means you have to find the area of both shapes and then compare them to see if Liam is correct if he is not correct say why. Hope this helps :)

3 0

4 years ago