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3 years ago

Algebra ll A cumulative exam answers ???

Mathematics

1 answer:

myrzilka [38]3 years ago

5 0

Answer: #1 3.14569 #GimmeMorePointsThenIGiveUSecondAnswer

Step-by-step explanation:

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OLga [1]

Answer:

x ≥ 10

Step-by-step explanation:

So put a filled-in dot on 10 and an arrow going to the right

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3 years ago

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C is the answer all you had to do was
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3 years ago

Find the solution of the separable equations a) the general solution: displaystyle dy/dt = t/[y^2] b) the unique solution to: dy

Xelga [282]

Answer:

(a) The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$ .

(b) The unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$ .

Step-by-step explanation:

(a)

The given differential equation is

$\frac{dy}{dt}=\frac{t}{y^2}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$y^2dy=tdt$

Integrate both sides.

$\int y^2dy=\int tdt$

$\frac{y^3}{3}=\frac{t^2}{2}+C$

The general solution of given differential equation is $\frac{y^3}{3}=\frac{t^2}{2}+C$ .

(b)

The given differential equation is

$\frac{dy}{dt}=\frac{1}{y}$

Use variable separable method, to solve the above differential equation.

Separate the variables.

$ydy=1dt$

Integrate both sides.

$\int ydy=\int 1dt$

$\frac{y^2}{2}=t+C$ ... (1)

It is given that y=1 at t=3. Substitute y=1 and t=3 in the above equation.

$\frac{(1)^2}{2}=(3)+C$

$\frac{1}{2}-3=C$

$-\frac{5}{2}=C$

Substitute $C=-\frac{5}{2}$ in equation (1).

$\frac{y^2}{2}=t-\frac{5}{2}$

Therefore the unique solution is $\frac{y^2}{2}=t-\frac{5}{2}$ .

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4 years ago

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The scale the speed of the car measured is:

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3 years ago

Please help, I’ll mark you as brainliest!!!!!!!!!!!!!

Sever21 [200]

Answer:

OOP BACK AT IT AGAIN.

The answer is B. 7 times 12 equals 84. 80 is the goal you are trying to meet.

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