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1 year ago

How many circles are there for each box

Mathematics

1 answer:

MaRussiya [10]1 year ago

7 0

what box? need further Intel

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Is it true that (x^2–3x+2)/(x –1) = x – 2, for all values of x? Justify your answer.

Nimfa-mama [501]

Answer:

x²-3x+2=0

x²-2x-x+2=0

(x²-2x)(-x+2)=0

x(x-2)-1(x-2)=0

(x-2)(x-1)=0

x-2=0 or x-1=0

x=2 or x=1

6 0

2 years ago

A spherical balloon is inflated with gas at a rate of 500 cubic centimeters per minute. Find the rates of change of the radius w

-BARSIC- [3]

Answer:

for r= 70 cm

the rate is 0.008cm/s

for r=95cm

the rate= 0.0044cm/s

Step-by-step explanation:

kindly find attached a rough sketch of the circle and a detailed solution of the problem.

Given data

rate dv/dt=500cm^3

r=70cm

r=95cm

4 0

2 years ago

Help me pls !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

defon

Answer:

for part A the answer is B and for part B I think the answer is B too.

3 0

2 years ago

the parabola y=x^2 is shifted up by 2 units and to the right by 3 units. what is the equation of the new porabola

Juli2301 [7.4K]

Up 2 units = y shifted up by 2:
y = x^2 + 2
to the right 3 units = x shifted right (+) by 3:
y + 3 = x^2 +2
new equation:
y = x^2 - 1

6 0

2 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.

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

Then

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

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago