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4 years ago

The diameter at the center of the tower is_ meters. The center of the tower is _meters above the ground

Mathematics

2 answers:

alexandr1967 [171]4 years ago

6 0

Answer:

think 5 meters

Step-by-step explanation:

Daniel [21]4 years ago

3 0

Answer:

The diameter at the center of the tower is 4 meters. The center of the tower is 8 meters above the ground.

Step-by-step explanation:

4x^2-y^2+16y-80=0

Completing squares in variable "y": Common factor -1:

4x^2-(y^2-16y)-80=0

4x^2-[(y-16/2)^2-(16/2)^2]-80=0

4x^2-[(y-8)^2-(8)^2]-80=0

4x^2-[(y-8)^2-64]-80=0

Eliminating the brackets:

4x^2-(y-8)^2+64-80=0

Adding like terms (constants):

4x^2-(y-8)^2-16=0

Adding 16 both sides of the equation:

4x^2-(y-8)^2-16+16=0+16

4x^2-(y-8)^2=16

Dividing all the terms by 16:

4x^2/16-(y-8)^2/16=16/16

Simplifying:

x^2/4-(y-8)^2/16=1

The hyperbola has the form:

(x-h)^2/a^2-(y-k)^2/b^2=1

Then:

h=0

k=8

a^2=4→sqrt(a^2)=sqrt(4)→a=2

The diameter (d) at the center of the tower is:

d=2a→d=2(2)→d=4 meters

The center of the tower is 8 (k) meters above the ground.

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3 1/5 x 4 1/2 ayuda por favor

Harman [31]

Answer:

14 2/5

Step-by-step explanation:

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3 years ago

Help, I need it by 9:00 am

olya-2409 [2.1K]

<h3>Answer:</h3>

Option B. because all rational numbers are integers.

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2 years ago

Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x

Serggg [28]

I assume $C$ has counterclockwise orientation when viewed from above.

By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

so we first compute the curl:

$\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k$

$\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k$

Then parameterize $S$ by

$\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k$

where the $z$ -component is obtained from

$1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$ .

Take the normal vector to $S$ to be

$\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k$

Then the line integral is equal in value to the surface integral,

$\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}$

6 0

3 years ago

Solve for w. 5w + 9z = 2z + 3w w = w equals negative StartFraction 7 Over 2 EndFraction z.Z w = w equals negative StartFraction

TEA [102]

Answer:

w = -7z/2

Step-by-step explanation:

We need to solve for w in terms of z.

The equation given is:

5w + 9z = 2z + 3w

Collect like terms:

5w - 3w = 2z - 9z

2w = -7z

w = -7z/2

5 0

3 years ago

Read 2 more answers

I need help plz help me

slega [8]

64 is the answer because it’s 8 for one And 8*8=64

3 0

3 years ago

Read 2 more answers