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3 years ago

F. How does sound propagate in air? Explain in short with a diagram.

Mathematics

1 answer:

anyanavicka [17]3 years ago

7 0

Answer:

Sound propagates through air as a longitudinal wave. The speed of sound is determined by the properties of the air, and not by the frequency or amplitude of the sound. Sound waves, as well as most other types of waves, can be described in terms of the following basic wave phenomena

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Which equation, when solved, results in a different value of x than the other three?

aev [14]

Solve each one to find that D is the answer

8 0

3 years ago

Read 2 more answers

Find the surface area of the composite figure.

ELEN [110]

Answer:

$644cm^2$

Step-by-step explanation:

First we'll work out the surface area of the pink figure.

That's the area of the two 5 by 6 rectangles on the top and bottom, plus the area of the two 5 by 20 and two 6 by 20 rectangles on the sides.

However, we note that the purple figure is blocking out a 6 by 12 section on the pink figure, so we'll need to subtract this.

The above works out to $2\times(30+100+120)-72=428 cm^2$ .

Then we'll work out the surface area of the purple figure.

This will be the area of the two 4 by 6 rectangles at the top and bottom, plus the area of the two 4 by 12 and one 6 by 12 rectangles on the sides. Note that there's only one 6 by 12 rectangle because the other face is joined to the pink figure, so it's blocked out.

That's $2\times(24+48)+72=216cm^2$ .

So the total surface area is $428+216=644cm^2$ .

6 0

1 year ago

A manufacturer produces bearings, but because of variability in the production process, not all of the bearings have the same di

Lena [83]

Answer:

Proportion of all bearings falls in the acceptable range = 0.9973 or 99.73% .

Step-by-step explanation:

We are given that the diameters have a normal distribution with a mean of 1.3 centimeters (cm) and a standard deviation of 0.01 cm i.e.;

Mean, $\mu$ = 1.3 cm and Standard deviation, $\sigma$ = 0.01 cm

Also, since distribution is normal;

Z = $\frac{X -\mu}{\sigma}$ ~ N(0,1)

Let X = range of diameters

So, P(1.27 < X < 1.33) = P(X < 1.33) - P(X <=1.27)

P(X < 1.33) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.33 -1.3}{0.01}$ ) = P(Z < 3) = 0.99865

P(X <= 1.27) = P( $\frac{X -\mu}{\sigma}$ < $\frac{1.27 -1.3}{0.01}$ ) = P(Z < -3) = 1 - P(Z < 3) = 1 - 0.99865

= 0.00135

P(1.27 < X < 1.33) = 0.99865 - 0.00135 = 0.9973 .

Therefore, proportion of all bearings that falls in this acceptable range is 99.73% .

7 0

3 years ago

What is the answer to this sum 5/6 of 120

ELEN [110]

100 is the correct answer

6 0

3 years ago

Prove that: (b²-c²/a)CosA+(c²-a²/b)CosB+(a²-b²/c)CosC = 0

IRISSAK [1]

Prove that:

$\:\:\sf\:\:\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C=0$

Proof: 

We know that, by Law of Cosines,

$\sf \cos A=\dfrac{b^2+c^2-a^2}{2bc}$
$\sf \cos B=\dfrac{c^2+a^2-b^2}{2ca}$
$\sf \cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Taking LHS

$\left(\dfrac{b^2-c^2}{a}\right)\cos A+\left(\dfrac{c^2-a^2}{b}\right)\cos B +\left(\dfrac{a^2-b^2}{c}\right)\cos C$

Substituting the value of cos A, cos B and cos C,

$\longmapsto\left(\dfrac{b^2-c^2}{a}\right)\left(\dfrac{b^2+c^2-a^2}{2bc}\right)+\left(\dfrac{c^2-a^2}{b}\right)\left(\dfrac{c^2+a^2-b^2}{2ca}\right)+\left(\dfrac{a^2-b^2}{c}\right)\left(\dfrac{a^2+b^2-c^2}{2ab}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2-a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2-b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2-c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^2-c^2)(b^2+c^2)-(b^2-c^2)(a^2)}{2abc}\right)+\left(\dfrac{(c^2-a^2)(c^2+a^2)-(c^2-a^2)(b^2)}{2abc}\right)+\left(\dfrac{(a^2-b^2)(a^2+b^2)-(a^2-b^2)(c^2)}{2abc}\right)$

$\longmapsto\left(\dfrac{(b^4-c^4)-(a^2b^2-a^2c^2)}{2abc}\right)+\left(\dfrac{(c^4-a^4)-(b^2c^2-a^2b^2)}{2abc}\right)+\left(\dfrac{(a^4-b^4)-(a^2c^2-b^2c^2)}{2abc}\right)$