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3 years ago

Analyze the diagram below and complete the instructions that follow.

Mathematics

1 answer:

Sladkaya [172]3 years ago

8 0

The large triangle is isosceles, so that the two smaller triangles that appear inside the large one have bases of 12/2, or 6.

Let's see how x and y are related:

sin 60 degrees = opp/hyp = x/y = sqrt(3) / 2

and cos 60 degrees = adj/hyp = 1/2 = 6 / y. Then y = 12 (which should be no surprise, since the large triangle is equilateral).

Then: x/y = sqrt(3) / 2, or

x sqrt(3) x
---- = ----------- = ------ , so that 2x = 12sqrt(3), or x=6sqrt(3).
y 2 12

In summary,
x = 6√3
y=12

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3 years ago

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stellarik [79]

Answer:

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Step-by-step explanation:

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3 years ago

A. How many ways can the letters of the word ALGORITHM be arranged in a row?

Tpy6a [65]

Answer:

a). The number if ways it can be arranged in a row= 362880 ways

b). 40320 ways

c). 5040 ways

Step-by-step explanation:

There are 9 alphabets in the word ALGORITHM

The number if ways it can be arranged in a row= 9!

The number if ways it can be arranged in a row= 9*8*7*6*5*4*3*2*1

The number if ways it can be arranged in a row= 362880 ways

if A and L must remain together (in order) as a unit, then we take it as 8 alphabets

= 8!

= 8*7*6*5*4*3*2*1

= 40320 ways

if the letters GOR must remain together (in order) as a unit, then we have 7 alphabets units remaining

= 7!

= 7*6*5*4*3*2*1

= 5040 ways

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3 years ago

A certain semiconductor device requires a tunneling probability of T = 10-5 for an electron tunneling through a rectangular barr

Goryan [66]

Answer:

Generally the barrier width is $a = 1.9322 *10^{-9} \ m$

Step-by-step explanation:

From the question we are told that

The tunneling probability required is $T = 1 * 10^{-5}$

The barrier height is $V_o = 0.4 eV$

The electron energy is $E = 0.08eV$

Generally the wave number is mathematically represented as

$k = \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }$

Here m is the mass of the electron with the value $m = 9.11 *10^{-31} \ kg$

h is is know as h-bar and the value is $\= h = 1.054*10^{-34} \ J \cdot s$

$k = \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }$

=> $k = 3.073582 *10^{9} \ m^{-1}$

Generally the tunneling probability is mathematically represented as

$T = 16 * \frac{E}{V_o } * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}$

$1.0 *10^{-5} = 16 * \frac{0.04}{0.4 } * [1 - \frac{0.04}{0.4} ] * e^{-2 * 3.0736 *10^{9} * a}$

=> $6.944*10^{-6}= e^{-2 * 3.0736 *10^{9} * a}$

Taking natural log of both sides

$ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}$

=> $-11.8776 = -2 * 3.0736 *10^{9} * a}$

=> $a = 1.9322 *10^{-9} \ m$

4 0

3 years ago