All categories

2 years ago

Find the area of circle B in term of ( pie )

Mathematics

1 answer:

natta225 [31]2 years ago

6 0

Answer:

Step-by-step explanation:

$1.5^2\pi =2.25\pi$

You might be interested in

How far is 10% of a 2,000 kilometer trip

Dmitry [639]

Answer:200

Step-by-step explanation:

2000x0.1=200

200

6 0

3 years ago

Read 2 more answers

Lisa drank 7/16 of a bottle of water during the soccer game. Julianne drank 2/3 of the water in a bottle that was the same size

frez [133]

Answer:

julianne drank by about .1 tenth

Step-by-step explanation:

5 0

2 years ago

PLEASE HELP<br> Find the value of y

Archy [21]

Answer:

Letter D! That's the answer

5 0

2 years ago

Read 2 more answers

Task: Population of a Species

Cloud [144]

Answer:

C(t) = 20t

Step-by-step explanation:

The graph of C(t) appears to be linear

The y intercept is 0

The slope is 200/10 = 20

y = mx+b where m is the slope and b is the y intercept

y = 20x+0

y = 20x

C(t) = 20t

4 0

2 years ago

Read 2 more answers

Find general solutions of the differential equation. Primes denote derivatives with respect toÂ x.

zepelin [54]

Answer:

$\mathbf{3x^2y^3+2xy^4=C}$

Step-by-step explanation:

From the differential equation given:

$6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0$

The equation above can be re-written as:

$6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0$

$(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0$

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

$\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial x}}} \ \ \text{at each point of R}$

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

$M(x,y) = 6xy^3 +2y^4\ and \ N(x,y) = 9x^2 y^2 +8xy^3$

So;

$\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3$

$\dfrac{\partial N}{\partial y }$

Let's Integrate $\dfrac{\partial F}{\partial x}= M(x,y)$ with respect to x

Then;

$F(x,y) = \int (6xy^3 +2y^4) \ dx$

$F(x,y) = 3x^2 y^3 +2xy^4 +g(y)$

Now, we will have to differentiate the above equation with respect to y and set $\dfrac{\partial F}{\partial x}= N(x,y)$ ; we have:

$\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0 \\ \\ g(y) = C_1$

Hence, $F(x,y) = 3x^2y^3 +2xy^4 +g(y) \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1$

Finally; the general solution to the equation is:

$\mathbf{3x^2y^3+2xy^4=C}$

5 0

2 years ago