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3 years ago

Is it possible to make a triangle with measure of 3mm 9mm and 5mm

Mathematics

1 answer:

tatyana61 [14]3 years ago

4 0

Answer: No

Step-by-step explanation: if you mean the measure of a is 3, the measure of b is 9 and the measure of c is 5, c being the hypotenuse then the answer is no because the the sides of a and b must be less than the measure of c. If 5 wasn’t the hypotenuse then the answer would be different.

(KEEP IN MIND THAT IF 9 IS THE HYPOTENUSE THEN THE ANSWER IS YES)

Hope this helps! :)

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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

My Notes A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 f

Sergio [31]

Answer:

6.667 ft/s

Step-by-step explanation:

Given

Height of street light = 15ft

Height of man = 6ft

Speed away from pole = 4ft/s

Let x represents the distance between the man and the pole, and Let y represents the distance between the tip of the man's shadow to the pole.

This forms a similar triangle (see attachment below).

From similar triangles, we have

(y - x)/6 = y/15 ----- Solve equation

15(y - x) = 6 * y

15y - 15x = 6y ---- Collect Like Terms

15y - 6y = 15x

9y = 15x --- divide through by 9

y = 15x/9

y = 5x/3 ---- differentiate with respect to time, t

dy/dt = 5/3 dx/dt

dx/dt represents rate of change of distance per time = 4ft/s

while dy/dt represents rate of movement of his shadow tips

dy/dt = 5/3 * 4

dy/dt = 20/3 = 6.667 ft/s

4 0

3 years ago

Read 2 more answers

Y - 3 = 2(x + 1)

Natalija [7]

Answer:

Step-by-step explanation:

6 0

3 years ago

A corporate bond has a coupon rate of 5.5 percent, a $1,000 face value, and matures three years from today. The corporation is i

melomori [17]

Answer:

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

Step-by-step explanation:

= (probability of entire face value paid*face value+probability of entire face value not paid*percent of face value paid*face value)/(1+discount rate)^years to maturity

probability of entire face value paid = 75%

face value = 1000

probability of entire face value not paid = 25%

percent of face value paid= 60%

discount rate = 15%

years to maturity = 3

$= \frac{\frac{75}{100}\times 1000 + \frac{25}{100} \times \frac{60}{100}\times 1000 }{(1+\frac{15}{100})^3 }$

$=\frac{0.75\times 1000 + 0.25\times 0.60 \times 1000}{(1+0.15)^3}$

$=\frac{750+0.25\times 0.60\times 1000}{1.15^3} \\\\=\frac{750+150}{1.520875} =\frac{900}{1.520875} \\\\=591.76$

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3 years ago

What type of quadrilateral has to the vertices A(3,6), B(3,3), C(6,3), and D(6,6)

prisoha [69]

a square as the distance between the lengths and widths are 3

8 0

3 years ago

Is it possible to make a triangle with measure of 3mm 9mm and 5mm​

Is it possible to make a triangle with measure of 3mm 9mm and 5mm