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poizon [28]
3 years ago
13

I need help with number 8

Mathematics
2 answers:
Alex Ar [27]3 years ago
7 0

Answer:

C

Step-by-step explanation:

zysi [14]3 years ago
5 0

Answer:

C. The circle is circumscribed about ΔABC

Explanation:

The circumscribed circle of a polygon is a circle that passes through all the vertices of the polygon.

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A human hair is approximately 40 microns (um) wide. If 1 um is equal to 10
Natasha2012 [34]

Answer:

D 40,000nm

Step-by-step explanation: 1nm/10^-9m * 10^-6/1um = 10^3 nm/um

Therefore in 40 microns there are:

40um * 10^3 nm/um = 40 * 10^3 nm = 40,000 nm

5 0
3 years ago
Given ABCD - WXYZ, Find the length of the segment AD, If Ab=12 WX=20 and WZ=8
hodyreva [135]

Answer:

8

Step-by-step explanation:

the length of segment AD would be 8 since segment AD and segment WZ are equal in length!

8 0
3 years ago
Please help me with the below question.
o-na [289]

Given that

\vec r(t) = \left\langle 4t e^{-t}, 8 \arctan(t), 8 e^t \right\rangle

we first differentiate with respect to t to get the tangent vector, \vec T(t) :

\vec T(t) = \dfrac{d\vec r}{dt} = \left\langle (4-4t) e^{-t}, \dfrac8{1+t^2}, 8e^t \right\rangle

At t = 0, the tangent vector is

\vec T(0) = \left\langle 4, 8, 8 \right\rangle

To get the <em>unit</em> tangent vector, multiply this by 1/(norm of tangent vector) :

\|\vec T(0)\| = \sqrt{4^2 + 8^2 + 8^2} = \sqrt{144} = 12

Then the unit tangent vector is

\dfrac{\vec T(0)}{\|\vec T(0)\|} = \left\langle \dfrac4{12}, \dfrac8{12}, \dfrac8{12}\right\rangle = \boxed{\left\langle\dfrac13, \dfrac23, \dfrac23\right\rangle}

7 0
2 years ago
Plz help I have an exam
RUDIKE [14]
Oh okie so I can see you in the first week or so
4 0
3 years ago
A 1.5-mm layer of paint is applied to one side of the following surface. Find the approximate volume of paint needed. Assume tha
Mariulka [41]

Answer:

V = 63π / 200  m^3

Step-by-step explanation:

Given:

- The function y = f(x) is revolved around the x-axis over the interval [1,6] to form a spherical surface:

                                 y = √(42*x - x^2)

- The surface is coated with paint with uniform layer thickness t = 1.5 mm

Find:

The volume of paint needed

Solution:

- Let f be a non-negative function with a continuous first derivative on the interval [1,6]. The Area of surface generated when y = f(x) is revolved around x-axis over the interval [1,6] is:

                           S = 2*\pi \int\limits^a_b { [f(x)*\sqrt{1 + f'(x)^2} }] \, dx

- The derivative of the function f'(x) is as follows:

                            f'(x) = \frac{21-x}{\sqrt{42x-x^2} }

- The square of derivative of f(x) is:

                            f'(x)^2 = \frac{(21-x)^2}{42x-x^2 }

- Now use the surface area formula:

                           S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2} *\sqrt{1 + \frac{(21-x)^2}{42x-x^2 } }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+(21-x)^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{42x-x^2+441-42x+x^2} }] \, dx\\\\S = 2*\pi \int\limits^6_1 { [\sqrt{441} }] \, dx\\S = 2*\pi \int\limits^6_1 { 21} \, dx\\\\S = 42*\pi \int\limits^6_1 { dx} \,\\\\S = 42*\pi [ 6 - 1 ]\\\\S = 42*5*\pi \\\\S = 210\pi

- The Volume of the pain coating is:

                           V = S*t

                           V = 210*π*3/2000

                          V = 63π / 200 m^3

8 0
4 years ago
Read 2 more answers
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