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3 years ago

What is 3/4 divided by 1/8

Mathematics

2 answers:

barxatty [35]3 years ago

5 0

Answer:

6 or 6/1

Step-by-step explanation:

Paul [167]3 years ago

3 0

Answer:

Step-by-step explanation:

keep 3/4 and flip 1/8

3/4x8/1=

cancel out the 4 and 8

3/1x2/1= 6/1= 6

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Three students, Alicia, Benjamin, and Caleb, are constructing a square inscribed in a circle with center at point C. Alicia draw

True [87]

Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

<h3>Inscribing a square</h3>

The steps involved in inscribing a square in a circle include;

A diameter of the circle is drawn.

A perpendicular bisector of the diameter is drawn using the method described as the perpendicular of the line sector. Also known as the diameter of the circle.

The resulting four points on the circle are the vertices of the inscribed square.

Alicia deductions were;

Draws two diameters and connects the points where the diameters intersect the circle, in order, around the circle

Benjamin's deductions;

The diameters must be perpendicular to each other. Then connect the points, in order, around the circle

Caleb's deduction;

No need to draw the second diameter. A triangle when inscribed in a semicircle is a right triangle, forms semicircles, one in each semicircle. Together the two triangles will make a square.

It can be concluded from their different postulations that Benjamin is correct because the diameter must be perpendicular to each other and the points connected around the circle to form a square.

Thus, Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

Learn more about an inscribed square here:

brainly.com/question/2458205

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6 0

2 years ago

The volume of this cone is 36π cubic units.

kap26 [50]

Answer:

108pi

Step-by-step explanation:

We do not need to know the height and radius, as the formula for the volume of a cone is just a third of the volume of a cylinder. So we have 36pi * 3 or 108pi.

4 0

3 years ago

Round 1.983 to the nearest hundredth

svet-max [94.6K]

Answer:

1.98

Step-by-step explanation:

Find the number in the hundredth place 8 and look one place to the right for the rounding digit 3. Round up if this number is greater than or equal to 5 and round down if it is less than 5.

4 0

3 years ago

HELP ASAP PLEASE 3 extra points if you get it right

-BARSIC- [3]

It would be the origin

8 0

3 years ago

Read 2 more answers

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

4 0

3 years ago