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dem82 [27]
3 years ago
11

Someone pls help me

Mathematics
1 answer:
evablogger [386]3 years ago
3 0

Answer:

a, f

Step-by-step explanation:

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The part of the sphere x2 + y2 + z2 = 16 that lies above the cone z = x2 + y2 . (Enter your answer as a comma-separated list of
statuscvo [17]

Answer:

(x,y,z)=(ucos(v), usin(v), \sqrt{16-u^{2})

where 0\leq u\leq 2\sqrt{2} and 0\leq v\leq 2\pi

Step-by-step explanation:

Equation of a cone is z=\sqrt{x^{2} +y^{2}}

Equation of a paraboloid is z=x^{2} +y^{2}

I have parametrised cone here. Please note that equation for cone in the question, is actually a paraboloid.

Imagine a sphere of radius 4, centered at origin and intersecting a cone also centered at origin and height along positive z-axis, given by the equations

x^{2} +y^{2} +z^{2} = 16\\z=\sqrt{x^{2} +y^{2}}

where z\geq x^{2} +y^{2}

Solving for these two equations, and substituting for z in the equation of sphere, we get a circle of radius 2\sqrt{2} units.

The equation of intersecting circle is:

x^{2} +y^{2}=8

Now, according to question, parametrizing this region of circle using parameters u and v.

Consider cylindrical co-ordinates: (r,θ,z)

In cylindrical co-ordinates

(x, y, z)= (r cos(θ),  r sin(θ), z)

x^{2} +y^{2}= r^{2}

Eliminating z, and changing (r, θ)=(u,v)

For cone: x=ucos(v)

y= usin(v)

z=\sqrt{16-u^{2}

or (x,y,z)=(ucos(v), usin(v), \sqrt{16-u^{2})

where 0\leq u\leq 2\sqrt{2} and 0\leq v\leq 2\pi

6 0
3 years ago
Evaluate (f-g)(-2)? precalc
castortr0y [4]
-2f+2g    is what I got


Hope the helps
8 0
3 years ago
#8 pleaseee help it’s due in 20 minssss
iris [78.8K]
The picture is very blurry I can’t seee
4 0
3 years ago
Mr. Volok buys 2.4 kg of sugar for his bakery.
belka [17]

Answer:

(a). 12 bags,

(b). 6 bags.

Step-by-step explanation:

We have been given that Mr. Volok buys 2.4 kg of sugar for his bakery.

(a). To find the number of 0.2 kg sugar bag that can Mr. Volak fill from 2.4 kg, we will divide 2.4 by 0.2.

\text{Number of 0.2 kg sugar bags}=\frac{2.4}{0.2}

Multiply numerator and denominator by 10:

\text{Number of 0.2 kg sugar bags}=\frac{24}{2}

\text{Number of 0.2 kg sugar bags}=12

Therefore, Mr. Volok can make 12 bags of 0.2 kg sugar in each.

(b). To find the number of 0.4 kg sugar bag that can Mr. Volak fill from 2.4 kg, we will divide 2.4 by 0.4.

\text{Number of 0.4 kg sugar bags}=\frac{2.4}{0.4}

Multiply numerator and denominator by 10:

\text{Number of 0.4 kg sugar bags}=\frac{24}{4}

\text{Number of 0.4 kg sugar bags}=6

Therefore, Mr. Volok can make 6 bags of 0.4 kg sugar in each.

4 0
4 years ago
Use the table of function rule, y=-10x+2
monitta
More information please?
6 0
4 years ago
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