All categories

3 years ago

Can someone please help me with this:)

Mathematics

1 answer:

madam [21]3 years ago

3 0

Answer: c

Step-by-step explanation:

You might be interested in

Soledads regular pay rate is $7.50 per hour. If she works more than 40 hours in one week, she receives 1.5 times

monitta

Answer:

Step-by-step explanation:

pay for 40 hours: 7.50 * 40= 300

pay for 1 hour of overtime pay: 1.5 * 7.5= 11.25 per hour

11.25 * 4= 45

45 + 300= 345

40 + 4= 44

8 0

3 years ago

If B =1-6w2 and A = 1+ w, find an expression that equals 2B – A in<br> standard form.

pychu [463]

Answer:

I am also still waiting for the answer, if you do find it PLEASE let me know!

Step-by-step explanation:

5 0

2 years ago

Through:(-1, 2) and (0,2)

Elis [28]

What are you asking for exactly ?

5 0

2 years ago

What is the answer I need the answer

scoray [572]

Answer:

The correct choice is A

Step-by-step explanation:

the definition of inverse is the opposite of something or different in direction so that would mean they want you to switch what they gave you so that would mean the answer is choice A

4 0

2 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

Can someone please help me with this:) ​

Can someone please help me with this:)