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3 years ago

The speed of sound in dry air at 20° C (68° F) is 1.236 × 103 km/h. What is the speed of sound written as an ordinary number?

Mathematics

1 answer:

defon3 years ago

7 0

Step-by-step explanation:

The speed of sound has nothing to do with "sea level" and sound pressure and without the medium air there is no speed of sound. In SI units with dry air at 20°C (68°F), the speed of sound c is 343 meters per second (m/s). This also equates to 1235 km/h, 767 mph, 1125 feet per second (ft/s), or 666 knots.

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This design shows several circles with the same center. The total radius of the design is 8 inches. The angle shown has a measur

Burka [1]

Answer:

11.33 in. to the nearest hundredth.

Step-by-step explanation:

The perimeter of the shaded area = length of the 2 straight lines + the length of the 2 arcs = 4 + length of the 2 arcs.

Calculate the length of the outer arc:

This equals (30 / 360) * perimeter of the largest circle

= 1/12 * 2 π * 8

= 4/3 π in.

The inner circle has a radius of 8 - 2 = 6 ins

so the length of the inner arc

= 1/12 * π * 2 * 6

= π in.

So the perimeter of the shaded region = 4 + 4/3 π + π

= 4 + 7π/3

= 11.33 in.

7 0

3 years ago

Read 2 more answers

The world's population has grown at an average rate of 1.9 percent per year since 1945. There were approximately 4 billion peopl

kkurt [141]

Answer: $f(t) = 4000000000( 1 .019)^t$

Step-by-step explanation:

Here, According to the question, The population of the world is 4 billion (approx).

Again According to the question, The world's population has grown at an average rate of 1.9 percent per year since 1945.

Therefore, after 1975 the growth rate will remains same.

Thus, the population of the world after t years,

$f(t) = 4000000000( 1+\frac{1.9}{100} )^t$ (By the formula $A = P(1+\frac{r}{100})^t$ )

⇒ $f(t) = 4000000000( 1.019 )^t$

Where 1.019 is the growth rate and 4 billion is the initial population.

6 0

3 years ago

On the basis of data collected during an experiment, a biologist found that the growth of a fruit fly population (Drosophila) wi

Ket [755]

Answer:

(a). 15

(b). 78

Step-by-step explanation:

Growth of the population of a fruit fly is modeled by

N(t) = $\frac{600}{1+39e^{-0.16t} }$

where t = number of days from the beginning of the experiment.

(a). For t = 0 [Initial population]

N(0) = $\frac{600}{1+39e^{-0.16\times 0} }$

= $\frac{600}{1+39}$

= $\frac{600}{40}$

= 15

Initial population of the fruit flies were 15.

(b).Population of the fruit fly colony on 11th day.

N(11) = $\frac{600}{1+39e^{-0.16\times 11} }$

= $\frac{600}{1+39e^{-1.76} }$

= $\frac{600}{1+39\times 0.172 }$

= $\frac{600}{1+6.71}$

= $\frac{600}{7.71}$

= 77.82

≈ 78

On 11th day number of fruit flies colony were 78.

3 0

4 years ago

20x^{2} +124x+24[/tex]

vova2212 [387]

Factorization is the process of breaking an integer down into its components or divisors, and further calculation can be defined as follows:

Given:

$20x^{2} +124x+24$

To find:

solve the given problem

Solution:

$\to \bold{20x^{2} +124x+24}\\\\\to \bold{4(5x^{2} +31x+6)}\\\\\to \bold{4(5x^{2} +31x+6)}\ \ \ \ \ \ \ \ \ \ \ 5\times 6= 30 \ \ and\ \ 30 \times 1= 30\\\\\to \bold{4(5x^{2} +(30+1)x+6)}\\\\\to \bold{4(5x^{2} +30x+1x+6)}\\\\\to \bold{4(5x(x +6)+1(x+6)}\\\\\to \bold{4((x +6) (5x+1)}\\\\$

Therefore, the final answer is " $\bold{4((x+6)(5x+1))}$ ".

Learn more:

brainly.com/question/10755001

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2 years ago

Need help with 5 and 6 will give branliest to the first to answer

DochEvi [55]

Answer:

Step-by-step explanation:

6 0

3 years ago

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