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3 years ago

What is the answer /6=40/60

Mathematics

1 answer:

ddd [48]3 years ago

7 0

There is no answer, the equation is false.

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Suppose that Dr. Vergara makes a free throw with probability 0.8 and, independently, Dr. Agor makes a free throw with probabilit

adelina 88 [10]

Answer:

The probability that Dr. Agor made his free throw is 40%.

Step-by-step explanation:

Given the fact that Dr. Agor usually converts his free throws with a probability of 0.4, to determine the percentage of probability that he will convert his free throw, the following calculation has to be performed:

0.4 x 100 x 1 = X

40 x 1 = X

40 = X

Therefore, Dr. Agor has a 40% chance of making his free throw.

5 0

3 years ago

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X + 2 2/1=10<br> HELPPPPPPP!!!!!!!!!!!!!!!!!

irakobra [83]

Answer:

x+4=10
x=6
Give brainlist

3 0

3 years ago

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

3 years ago

Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz help

Sindrei [870]

Answer:

Step-by-step explanation:

Since both angles together need to equal 180, set up your formula as

(x+30)+(x+10)=180

2x+40=180

3 0

3 years ago

Solve. 7(2y-10)=14. What is the awnser

Sergeu [11.5K]

Answer:

y = 6

Step-by-step explanation:

3 0

3 years ago

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