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3 years ago

Prove the following statement (1-sinA)×sin²A/(1+cosA)×cos²A=1-cosA/1+sinA

Mathematics

1 answer:

shepuryov [24]3 years ago

4 0

Answer:

(1-sinA) * Sin²A

(1+cosA) * Cos²A

(1-sinA) * (1-Cos²A)

(1+cosA) * (1-Sin²A)

(1-sinA) * (1-CosA)(1+CosA)

(1+cosA) * (1-SinA) (1+SinA)

(1-CosA)

(1+SinA)

proved

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A large container has a maximum capacity of 64 ounces. The container is filled with 8 ounces less than its maximum capacity. Ans

son4ous [18]

Answer:

The container is filled to 87.5% of the maximum capacity.

Step-by-step explanation:

Given

Maximum capacity of container = m = 64 ounces

Filled capacity=> as it is 8 less than maximum capacity = 64-8 = 56 ounces

The container is 56 ounces full.

To find the percentage:

$percentage = \frac{56}{64}*100\\= 0.875*100\\=87.5\%$

Hence,

The container is filled to 87.5% of the maximum capacity.

3 0

3 years ago

Adult tickets for the movie theater cost $9. Child tickets cost $5.25. Write an expression for the total cost of a group of adul

sammy [17]

Answer:

Step-by-step explanation:

9a + 5.25c

5 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$